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COMBINATORIAL DICHOTOMIES IN SET THEORY

STEVO TODORCEVIC

In this article we give an overview of a line of research in set theory that hasreached a level of maturity and which, in our opinion, merits its being exposedto a more general audience. This line of research is concentrated on finding towhich extent compactness fails at the level of first uncountable cardinal and towhich extent it could be recovered at the level of some other not so large cardinal,the most interesting being of course the level of the second uncountable cardinal.While this is of great interest to set theorists, one of the main forces behind thisline of research stems from its applicability to other areas of mathematics whereone encounters structures that are not necessarily countable, and therefore one islikely to encounter problems of this kind. We have split this overview into threeparts each presenting a set theoretic combinatorial principle imposing a degree ofcompactness at some small cardinal. The three principles are natural dichotomiesabout chromatic numbers for graphs and about ideals of countable subsets of someindex set. They are all relatively easy to state and apply and are therefore accessibleto mathematicians working in areas outside of set theory. Each of these threedichotomies has its axiomatic as well as its mathematical side and we went intosome effort to show the close relationship between these. For example, we havetried to show that an abstract analysis of one of these three set theoretic principlescan sometimes lead us to results that do not require additional axioms at all butwhich could have been otherwise difficult to discover directly. We have also tried toselect examples with as broad range as possible but their choices could still reflecta personal taste. We have therefore included an extensive reference list where thereader can find a more complete view on this area of current research in set theory.Finally we mention that while this article is meant for a larger audience which istypically interested in a general overview, we have tried to make the article alsointeresting to experts working in this area by including some of the technicalitiesespecially if they appear to us as important tools in this area. For the same reasonwe will be pointing out a number of possible directions for further research.

Our terminology and notation follows that of standard texts of set theory (see,for example, [53] and [67]). Recall that the Singular Cardinals Hypothesis, SCH, isthe statement

(∀θ) θcf θ = θ+ · 2cf θ.

It is a strong structural assumption about the universe of sets as it answers allquestions about the arithmetic of infinite cardinals. It is for this reason one of themost studied such hypotheses of set theory especially in the context of coveringproperties of inner models of set theory. Another set of principles which is evenmore relevant from this point of view are the square principles such as 2κ and2(κ). Before we introduce these principles, recall that Cα (α < θ) is a C-sequenceif Cα is a closed and unbounded subset of α for all α < θ. These combinatorial

Date: Bern, July 3, 4, and 5, 2008.1

2 STEVO TODORCEVIC

principles 2κ and 2(κ) assert the existence of C-sequences Cα (α < θ) for θ = κ+

and θ = κ, respectively, with some extra properties most important of which is thecoherence property stating that

Cα = Cβ ∩ α whenever α is a limit point of Cβ .

For Cα (α < κ+) to form a 2κ-sequence the sets Cα have to be moreover all oforder type ≤ κ. On the other hand, for Cα (α < θ) to form a 2(θ)-sequence it hasto be moreover nontrivial in the sense that there is no closed and unbounded subsetC of θ such that Cα = C ∩ α for every limit point α of C. Note that 2(ω1) andin fact 2ω holds1 but the existence of such sequences on any of the higher regularcardinals is not a consequence of the usual axioms of set theory. It is also known(see [88]) that these principles hold in all currently known core models of set theoryand that the global failures of these principles is of a considerable large cardinalstrength. In particular, it is known that the failure of 2κ for strong limit singularκ or the simultaneous failure of the 2(θ)-principles have some important structuralconsequences most interesting of which is the principle of Projective Determinacy,PD, stating that all projective games are determined (see [99], [55], [89]).

Part 1. Combinatorial Compactness

We start Part I with a well known large cardinal axiom that has played a crucialrole in the development of set theory. Its influence on the cardinal arithmeticat singular cardinals and its influence on square sequences and consequently itsnonrelativization to any of the current core models of set theory form a basis ofthe interest in this axiom. We give a precise list of these consequences since weintend to compare them with the corresponding consequences of combinatorial settheoretic principles that we introduce in later parts of this article.

Definition 0.1 ([27]). An infinite cardinal κ is compact2 if every set S of sentencesof Lκκ that has no model contains a subset S0 of cardinality < κ without a model.3

Remark 0.2. Since ω is a compact cardinal, in what follows compact cardinal isusually required to be uncountable.

Here are some sample results that use this notion and which range over importantthemes of research of modern set theory. They serve as prototype themes for manyresults to be discussed in later sections of this article.

Theorem 0.3 ([97]). Let κ0 be the first uncountable compact cardinal. Then(1) (∀θ ≥ κ0) θcf θ = θ+ · 2cf θ.4

(2) κ fails for every κ ≥ κ0.

Theorem 0.4 ([89]). If there is a compact cardinal then all projective games aredetermined.

1In hindsight one can say that 2(ω1) and 2ω are directly responsible for the ‘incompactness at

the level of the first uncountable cardinal’ alluded to above at the beginning of this Introduction

(see [128] for more details).2Or strongly compact in the more recent literature.3Recall that in sentences of Lκλ we allow conjucntions and disjunctions of length < κ and

strings of quantifiers of length < λ.4 The statement (∀θ) θcf θ = θ+ · 2cf θ is the well-known Singular Cardinals Hypothesis, SCH,

So this is saying that the Singular Cardinals Hypothesis holds above the first uncountable compact

cardinal.

COMBINATORIAL DICHOTOMIES IN SET THEORY 3

Question 0.5. Is the compactness of the logic Lω1ω as strong? More precisely, doesthe existence of an (uncountable) cardinal κ for which the logic Lω1ω is κ-compact5

has consequences as these appearing in Theorems 0.3 and 0.4 above?

We shall see below that the answer is positive but we shall be particularly interestedin finding an optimal answer to the following variant of this question.

Question 0.6. What are the weak forms of compactness of this sort with as strongstructural consequences on the universe of set theory?

1. Compactness Principles for Graphs

Recall that a graph is a structure of the form G = (X,E), where X is a set,called a set of vertices of G, and where E is a symmetric irreflexive binary relationof X called the set of edges of G.

Example 1.1. The complete graph on the set of vertices Z is the graph KZ =(Z,Z2 \ 4) and the discrete graph with vertex-set Z is the graph DZ = (Z, ∅).

Given two graphs G = (X,E) and H = (Y, F ), a mapping f : X → Y is a homo-morphism from G into H whenever

(x, y) ∈ E implies (f(x), f(y)) ∈ Ffor all x, y ∈ X. We use the notation G ≤ H to denote the fact that there isa homomorphism from G to H. If the vertex sets X and Y are equipped withtopologies and measures then we use the notation G ≤C H, G ≤B H or G ≤L H torecord the fact that there is a homomorphism from G into H which is continuous,Borel or Baire measurable, or Lebesgue measurable, respectively.

Definition 1.2. The chromatic number of a graph G is defined by

Chr(G) = min|Z| : G ≤ KZ.When the homomorphisms into KZ are to be witnessed by continuous, Borel, Baire,or Lebesgue measurable maps, we use the notations ChrC(G), ChrB(G), ChrL(G),respectively, for the corresponding variations on the chromatic number.

Remark 1.3. Thus, a graph G has countable chromatic number, or is countablychromatic, if its set of vertices can be (colored) covered by countably many (colors)G-discrete sets, i.e., sets of vertices which contain no edges.

Definition 1.4. Given a class H of graphs, let κH be the minimal cardinal κ(provided it exists) such that every graph G ∈ H with

Chr(G) > ℵ0

has a subgraph G0 of cardinality < κ such that

Chr(G0) > ℵ0.

Remark 1.5. Clearly, for every class H of graphs,

κH ≤ the first uncountable compact cardinal.

Problem 1.6. Classify the classes of graphsH that have small compactness numberκH. In particular, classify the of classes of graphs H such that κH = ℵ2.

5Or in other words, which has the property that every set of sentences of Lω1ω which has no

model has a subset of cardinality < κ without a model.

4 STEVO TODORCEVIC

Note that ℵ2 is indeed the first possible nontrivial such a value for κH and so weare naturally led to the following question.

Question 1.7. Given a class H0 of graphs with κH0 equal to the minimal possiblevalue, can one find any interesting structure of the universe of sets above thiscardinal?

2. Classes of Graphs with Small Compactness Numbers

Example 2.1. Let G be the class of of all graphs. Not much is known about thecorresponding compactness number κG .

Theorem 2.2 ([26]). κG ≥ iωSketch of Proof: Let γ be an ordinal, and k ≥ 2 an integer. We define the graph([γ]k,Shift) by setting (α0, . . . , αk−1< , β0, . . . , βk−1<) ∈Shift if and only if

α0 < α1 = β0 < α2 = β1 < · · · < αk−1 = βk−2 < βk−1.

Claim 1. Chr([γ]k,Shift) ≤ ℵ0 whenever |γ| ≤ ik−1.

Proof. Induction on k but for the stronger statement Chr(γk,Shift

)≤ ℵ0.

Claim 2. Chr([γ]k,Shift) > ℵ0 whenever γ ≥ i+k−1.

Proof. Use i+k−1 → (k + 1)kℵ0

.

Remark 2.3. The construction makes perfect sense for finite ordinals as well.Moreover, the shift graph can be associated to any hypergraph in place of thecomplete hypergraph (γ, [γ]k). Let us explain this in case k = 2, i.e., in case ofa directed or an undirected graph. Thus, given a graph G = (X,E) and a totalordering < of its vertex set X, let

Shift(G) = (VShift, EShift)

be the graph whose vertex set VShift is equal to the set E of edges of G and where weput ((x, y), (x′, y′)) in the edge relation EShift of Shift(G) whenever x < y = x′ < y′.Then we have the following fact proved in a similar manner as above.

Theorem 2.4 ([26]). Chr(G) > 2κ implies Chr(Shift(G)) > κ.

3. Dilworth’s Theorem and Chromatic Numbers of IncomparabilityGraphs

In this section we investigate the class HD of graphs whose vertex sets are posets(or, more generally, quasi-ordered sets) and whose edges are pairs of incompara-ble points. The following famous result of Dilworth is giving us the compactnessnumber for families of k-chromatic such graphs, where k is some fixed finite number.

Theorem 3.1 ([21]). For a finite integer k ≥ 1, a poset P can be decomposed into≤ k chains iff every subposet P0 ⊆ P of cardinality ≤ k+ 1 can be decomposed into≤ k chains.6

It is therefore natural to investigate the compactness number of the family of allcountably chromatic incomparability graphs on posets.

6In other words, a poset can be covered by ≤ k chains iff it contains no antichain of size k+ 1.


Definition 3.2. Let κD be the minimal cardinal κ (if it exists) with the propertythat a partially ordered set P can be decomposed into countably many chains ifevery subordering P0 ⊆ P of size < κ can be decomposed into countably manychains.

Conjecture 3.3 (Galvin’s Conjecture). ℵ2 is a possible value of κD.

While Galvin’s conjecture is still an open problem the following result verifies itscorrectness for an important special case.

Theorem 3.4 ([47]). Galvin’s Conjecture holds in the class of Borel quasi-orderedsets defined on Polish spaces.7

4. Rado’s Theorem and Chromatic Numbers of Interval Graphs

In this section we investigate a special kind of graphs, the class HR of graphs ofthe form (I, E), where I is some family of nonempty intervals, or more generally,convex sets of some linearly ordered set L and where two intervals form an edge iftheir intersection is nonempty. The following result of Richard Rado computes thecompactness numbers for each of the families (I, E) ∈ HR : Chr(I, E) ≤ k forsome finite number k.

Theorem 4.1 ([83]). For an integer k ≥ 1, a family F of intervals of some linearlyordered set (L,<L) can be decomposed into ≤ k subfamilies of pairwise disjointintervals if this is true about any subfamily F0 ⊆ F of size ≤ k + 1.

It is therefore quite natural to investigate the compactness number of countablychromatic intersection graphs on intervals.

Definition 4.2. Let κR be the minimal cardinal κ (if it exists) with the propertythat a family F of intervals of some linearly ordered set (L,<L) can be decomposedinto countably many subfamilies of pairwise disjoint intervals if this is true aboutevery subfamily F0 ⊆ F of cardinality < κ.

Conjecture 4.3 ([84]). [Rado’s Conjecture] κR = ℵ2.

Theorem 4.4 ([114]). Rado’s Conjecture is equivalent to the statement that anarbitrary tree T can be decomposed into countably many antichains if and only ifevery subtree T0 of T of cardinality at most ℵ1 allows such a decomposition.

Corollary 4.5. κR ≤ κD, and so in particular, Rado’s Conjecture is a consequenceof Galvin’s Conjecture.

Proof. Consider a tree T = (T,≤T ) and pick a well-ordering <w of its domain T.Define another ordering ≤P on T by letting s ≤P t if and only if s = t or s andt are ≤T -incomparable and u(s) <w u(t), where u(s) and u(t) are the immediatesuccessors of the node v = s∧t extended by s and t, respectively. Then P = (T,≤T )is a partially ordered set living on the same domain T such that

(∀C ⊆ T ) C is a chain of P iff C is an antichain of T .So if the tree T cannot be covered by countably many antichains, the correspondingposet P is not the union of countably many chains, so there is T0 ⊆ T of size < κDsuch that P T0 and T T have the same corresponding properties. This showsthat κD ≥ κR.

7More precisely, if a Borel quasi-ordered set contains no perfect antichain then it can bedecomposed into countably many Borel chains.

6 STEVO TODORCEVIC

Remark 4.6. In fact in can be shown that Galvin’s Conjecture for the class offinite-dimensional8 posets is equivalent to Rado’s Conjecture.

Problem 4.7. Is κR a regular cardinal? If not, can it have countable cofinality?

Theorem 4.8 ([103]). The equality κR = ℵ2 is consistent relative to the consistencyof the existence of a compact uncountable cardinal.

We repeat that the consistency of Galvin’s Conjecture is still an open problem.The formulation of Rado’s Conjecture in terms of trees is a key step behind

basically all applications of this principle that appeared afterwards. The reason forthis perhaps lies in the fact that trees that are not countably chromatic 9 behavemuch like the first uncountable ordinal ω1. The following two facts are illustrationsof this phenomenon.

Lemma 4.9 ([101]). [Pressing-Down Lemma for Trees] For every non-special treeT and every regressive 10 mapping f : T → T there is a non-special subtree S of Ton which f is constant.

Lemma 4.10 ([101]). [∆-System Lemma for Trees] For every family Ft (t ∈ T ) offinite sets indexed by a non-special tree T there is a finite set R and a non-specialsubtree U of T such that Fs∩Ft = R whenever s, t ∈ U are chosen such that s <T t.

5. The Set-Theoretic Universe Above Rado’s Cardinal

Theorem 5.1 ([114]). The Singular Cardinals Hypothesis holds above κR. Moreprecisely,

(1) θλ = θ for regular θ, λ such that θ ≥ (κR)λ.(2) θcf θ = θ+ · 2cf θ for every θ ≥ (κR)ℵ0 .

Proof Sketch. The essential difficulty is in proving the equality θℵ0 = θ by inductionon regular cardinals θ ≥ (κR)ℵ0 . By Silver’s theorem (see [92]) we can afford toconcentrate on the case

θ = κ+ for some κ ≥ (κR)ℵ0 with cf(κ) = ω.

Choose a strictly increasing converging sequence κi κ consisting of regular car-dinals ≥ (κR)ℵ0 . For a, b ∈

∏i<ω κi set:

a <∗ b iff (∀∞i) a(i) < b(i), a ≤∗ b iff (∀∞i) a(i) ≤ b(i),a < b iff (∀i) a(i) < b(i), a ≤ b iff (∀i) a(i) ≤ b(i).

Choose A = aξ : ξ < κ+ ⊆∏i<ω κi such that

(1) ξ < η implies aξ ≤∗ aη,(2) ∀ξ(∃η > ξ)aξ <∗ aη,(3) A is closed under finite changes of its elements.

Choose τ : [κ+]2 → ω such that

8Recall, that the dimension of a poset (P,≤P ) is the minimal cardinality of a family ≤i (i ∈ I)of linear orderings of the set P such that ≤P=

⋂i∈I ≤i .

9More precisely, trees whose comparability graphs are not countably chromatic, or in otherwords, trees that cannot be decomposed into countably many antichains. Such trees are knownin the literature under the name of non-special trees.

10We say that a mapping f : T → T defined on a tree T is regressive if f(t) <T t for allnon-minimal t ∈ T.


(4) τ(α, γ) ≤ maxτ(α, β), τ(β, γ) whenever α < β < γ,(5) for all α < κ+ and n < ω, the set ξ < α : τ(ξ, α) ≤ n has size ≤ κn.

For a subset t of A, let sup(t) ∈∏i<ω κi be defined by sup(t)(i) = supa∈t a(i).

For α < κ+ and i < ω, set

Ai(α) = aξ : ξ < α & τ(ξ, α) ≤ iand

A∗i (α) =

sup(t) : t ∈ [Ai(α)]≤ℵ0.

By the inductive hypothesis |A∗i (α)| ≤ κi for all α < θ = κ+ and i < ω. Let

A∗ =⋃

α<κ+

⋃i<ω

A∗i (α).

Then |A∗| ≤ θ.Let TA be the set of all <-increasing countable sequences t = tξ : ξ ≤ µ of

successor length, such that

supξ<λ

tξ /∈ A∗ for all limit ordinals λ ≤ µ.

We consider TA as a tree ordered by end-extension and choose a lexicographicalordering on TA and let LA be the corresponding linear ordering. Finally, let

FA = s ∈ TA : t v s : t ∈ TA ,a family of intervals of LA.

Claim 1. Every subfamily of FA of cardinality ≤ κ can be decomposed into count-ably many subfamilies of pairwise disjoint intervals.

Claim 2. If FA itself can be decomposed into countably many subfamilies of pair-wise disjoint intervals there is a Namba subtree11 W ⊆

⋃k<ω

∏i<k κi such that the

set [W ] of all of its infinite branches is included in A∗. So, in particular, κℵ0 ≤ κ+.

These claims complete the proof of Theorem 5.1.

Recall that a C-sequence on an ordinal θ is simply a sequence Cα (α < θ) ofsets, where for each α < θ, the set Cα is closed and unbounded in α. We say thata C-sequence Cα (α < θ) is coherent if

Cα = Cβ ∩ α whenever α is a limit point of Cβ .

Clearly, the sequence Cα = α (α < θ) is coherent so we are interested in less trivialexamples. Thus, we say that a C-sequence Cα (α < θ) is trivial if there is a closedand unbounded set C ⊆ θ such that

Cα = C ∩ α whenever α is a limit point of C.

Let (θ) denote the statement that θ supports a nontrivial coherent C-sequenceCα (α < θ). Note that if θ = κ+ then every κ-sequence in the sense of [54] is a(θ)-sequence but in general (κ+) does not imply κ.

Theorem 5.2 ([114]). (θ) fails for all ordinals θ of cofinality ≥ κR.

11Here, we consider T =⋃k<ω

∏i<k κi as a tree ordered by extension. A subtree S of T

is a Namba subtree if it is downwards closed and if it preserves the degrees of nodes, i.e., if|ξ < κi : taξ ∈ S| = κi for every t ∈ S of length i.

8 STEVO TODORCEVIC

Sketch of Proof: Note that it suffices to prove this for regular cardinals θ ≥ κR.Towards a contradiction suppose we have a nontrivial coherent sequence Cα (α < θ).Consider the tree T of all countable closed subsets A of θ such that

sup(A ∩ Cα) < α for all α ∈ Awith the end-extension as the tree-ordering. Then T is a tree that cannot bedecomposed into countably many antichains though for all δ < θ, the subtreeT δ = A ∈ T : max(A) < δ does have such a decomposition. To see this, definef : T δ → T δ, by

f(A) = A ∩ (ξ + 1) where ξ = max(A ∩ Cδ.)Then f is a regressive mapping with the property that for every B ∈ T δ, thepre-image f−1(B) can be decomposed into countably many antichains.

This shows that the assumption κR < ∞ that there is a Rado cardinal hasconsiderable large-cardinal strength. Using this and some deep work about innermodels of set theory and determinacy one can state the following kind of directconsequences.

Theorem 5.3 ([55]). If there is a Rado cardinal then projective determinacy holds.

Corollary 5.4 ([55]). RC implies PD.

We finish this section with another direct consequences of the existence of Rado’scardinal. It can be seen as a corollary of the proof of Theorem 8.18 below.

Theorem 5.5. θ → (ω1)<ωω1,ωfor every regular θ ≥ κR.12

6. Rado’s Conjecture and the Continuum

Theorem 6.1 ([114]). Rado’s Conjecture implies c ≤ ℵ2.13

Proof. Fix an e : [ω2]2 → ω1 such that e(α, γ) 6= e(β, γ) for α < β < γ < ω2.Let Ce be the collection of all A ∈ [ω2]ℵ0 such that(1) A ∩ ω1 ∈ ω1

(2) α ∈ A ⇐⇒ α+ 1 ∈ A(3) α < β in A⇒ e(α, β) ∈ A(4) ν = e(α, β) & ν, β ∈ A⇒ α ∈ A.

Define

Se = A ∈ Ce : (∀ν < ω1) (∀α < ω2)A 6= ξ < α : e(ξ, α) < ν .

Note 6.2. If A $ B in Ce and A ∩ ω1 = B ∩ ω1 then A /∈ Se.

Note 6.3. Se ∩ [Y ]ℵ0 is not stationary in [Y ]ℵ0 for every uncountable Y ⊆ ω2.

Let Te be the tree of all countable continuous chains of elements of Se whoseunions are also elements of Se. Let Le be the linear ordering obtained by lexico-graphically ordering Te. For t ∈ Te, let

It = s ∈ Te : t v s ,

12Recall that θ → (ω1)<ωω1,ωmeans that for every mapping F : θ<ω → ω1 there exists an

uncountable set X ⊆ θ such that |f ′′X<ω | ≤ ℵ0.13It is known that Rado’s Conjecture does not decide whether c = ℵ1 or c = ℵ2 (see [103]).


a convex subset of the linearly ordered set Le. Let

Fe = It : t ∈ Te ,be a family of convex subsets of Le. For δ < ω2, let

Fδe =It ∈ Fe :

⋃t ⊆ δ

.

Claim 1. For each δ < ω2 the family Fδe can be decomposed into countably manysubfamilies consisting of pairwise disjoint intervals. So in particular, every subfam-ily of Fe of cardinality < ℵ2 can be decomposed into countably many subfamiliesconsisting of pairwise disjoint intervals.

Claim 2. If Fe itself can be decomposed into countably many subfamilies consistingof pairwise disjoint intervals, then R ⊆ L[e], so in particular c ≤ ℵ2.

Proof. Note that if R * L[e] then the set Se is a stationary subset of [ω2]ℵ0 .

This completes the proof of Theorem 6.1.

Theorem 6.4 ([114]). Rado’s Conjecture implies that every transitive model of asufficiently large fragment of ZFC which correctly computes ℵ2 must contain all thereals.

Proof. Suppose R * M . Then S = [ω2]ℵ0 \M is a stationary subset of [ω2]ℵ0 (see[10] and [44]). Choose f : [ω2]2 → ω1 such that f ∈ M and f(α, γ) 6= f(β, γ)whenever α < β < γ < ω2. Then S ∩ Ce ⊆ Se. Then Se is stationary and thereforeRado’s Conjecture is false.

7. Rado’s Conjecture and Chang’s Conjecture

Recall that Chang’s Conjecture, CC, is the statement that every structure

〈A,U, . . . 〉of countable signature where A has cardinality ℵ2 and where U is a distinguishedunary predicate of cardinality ℵ1 has an elementary substructure

〈B,B ∩ U, . . . 〉with |B| = ℵ1 and |B ∩ U | = ℵ0. This was one of the first combinatorial modeltheoretic statements on the level of ω2 with a considerable large cardinal strength.In this section we give some explanations of the following result which points outthe connection between RC and CC.

Theorem 7.1 ([114]). Rado’s Conjecture implies Chang’s Conjecture.

Since the proof of this result has ideas of general interest, we give some details.

Definition 7.2. The Game Gω(ω2, ω1) is played as follows

I f0 : ω2 → ω1 f1 : ω2 → ω1 · · ·II δ0 δ1 · · ·

where δi < ω2 for all i < ω. Player II wins f0, δ0, f1, δ1, . . . if

α < ω2 : fn(α) < supi<ω

δi for all n < ω

is an unbounded subset of ω2; otherwise I wins.

10 STEVO TODORCEVIC

Theorem 7.3. Rado’s Conjecture implies that II has a winning strategy in thegame Gω(ω2, ω1).

Proof. For A ∈ [ω1 ∪ ωω21 ]ℵ0 , let

DA = α < ω2 : (∀f ∈ A)f(α) ∈ A.

LetSG = A ∈ [ω1 ∪ ωω2

1 ]ℵ0 : A ∩ ω1 ∈ ω1 & supDA < ω2.

Claim 1. II has a winning strategy in Gω(ω2, ω1) if and only if SG is not a sta-tionary subset of [ω1 ∪ ωω2

1 ]ℵ0 .

Definition 7.4. For A,B ∈ SG we say B strongly includes A if

A ⊆ B and A ∩ ω1 < B ∩ ω1.

Let TG be the tree of all countable continuous strongly increasing chains t ofelements of SG such that

⋃t ∈ SG.

Let LG be the linearly ordered set obtained by lexicographically ordering thetree TG. For t ∈ TG, let

It = s ∈ TG : t v s,a convex set (interval) of the linearly ordered set LG. Let

FG = It : t ∈ TG .

Claim 1. Every subfamily F0 of FG of cardinality < ℵ2 can be decomposed intocountably many subfamilies of pairwise disjoint intervals.

Claim 2. If FG itself can be decomposed into countably many subfamilies of pair-wise disjoint intervals then SG is nonstationary and therefore II has a winningstrategy in the game Gω(ω2, ω1).

This completes the proof of Theorem 7.3.

Lemma 7.5. If II has a winning strategy in Gω(ω2, ω1) then for every countable

M ≺ (Hθ,∈, <) ,

for θ a sufficiently large regular cardinal, there is

M∗ ≺ (Hθ,∈, <)

such that M ⊆M∗, M∗ ∩ ω1 = M ∩ ω1, and M ∩ ω2 6= M∗ ∩ ω2.

Proof. Take M∗ = Sk (M ∪ α) for α ∈ DM∩(ω1∪ωω21 ) with α > sup(M∩ω2).14

Note that this finishes the proof of Theorem 7.1. The conclusion of Lemma 7.5is a strong form of Chang’s Conjecture which we denote by CC∗. This way ofstrengthening CC showed up first during an analysis of conditions that guaranteethe semi-properness of the standard Namba forcing which changes the cofinality ofω2 to ω (see [90]; pp. 355-400). The difference between CC and its strong formCC∗ can be seen from the following result.

Theorem 7.6 ([114]). CC∗ implies c ≤ ℵ2.

14We are using here the functor A 7→ DA from the proof of Theorem 7.3 above.


Recall that CC has no influence on the cardinality of the continuum since it ispreserved by forcing extensions over posets satisfying the countable chain condition.

Since the proof of Theorem 7.6 involves interesting ideas we give its sketch. Firstof all we note the following fact.

Lemma 7.7 ([114]). CC∗ implies that for every stationary S ⊆ [ω2]ℵ0 there existsuncountable α < ω2 such that S ∩ [α]ℵ0 is stationary in [α]ℵ0 .

Proof. Suppose S∩ [α]ℵ0 is not stationary whenever ω1 ≤ α < ω2, and for each suchα choose one-to-one eα : α → ω1 such that Aν(α) = ξ < α : eα(ξ) < ν 6∈ S forall non-zero limit ordinals ν < ω1. Since S is stationary, for every sufficiently largeregular cardinal θ, we can choose a countable M ≺ (Hθ,∈, <) such that M∩ω2 ∈ S.Let M∗ ≺ (Hθ,∈, <) be an ω2-end-extension of M guaranteed by CC∗ and letα = min((M∗ ∩ ω2) \M). Then Aν(α) = M ∩ ω2 ∈ S for ν = M ∩ ω1 = M∗ ∩ ω1,a contradiction.

The conclusion of Lemma 7.7 is known in the literature under the name of WeakReflection Principle for ω2, WRP(ω2).15 Thus we have established the following

RC⇒ CC∗ ⇒WRP(ω2).

So it remains to establish the following.

Theorem 7.8 ([105]). WRP(ω2) implies c ≤ ℵ2.

Proof. Since every closed and unbounded subset of [ω2]ℵ0 has cardinality at leastcontinuum (see [10]), if c > ℵ2, the set S = [ω2]ℵ0 \ Aν(α) : α < ω2, ν < ω1 isstationary. But clearly S ∩ [α]ℵ0 is not stationary in [α]ℵ0 for all α < ω2.

In [91] this result was extended to all other regular cardinals θ in place of ω216

which reiterates the following natural question.

Question 7.9 ([114]). Does RC imply that WRP(θ) for all cardinals θ ≥ ω2?

It turns out that CC∗ is the ω2-case of a general principle isolated in [23] as oneof the equivalents of the well-known assertion that forcing notions that preservestationary subsets of ω1 are semi-proper, a principle that has played an importantrole in the original proof of the consistency of Martin’s Maximum in [39].

Definition 7.10. For λ ≥ ω2, let CC∗(λ) be the statement that for an arbitraryregular cardinal θ > λ, a well-ordering <w of Hθ, a set a ∈ [λ]ω1 , and a countableM ≺ (Hθ,∈, <w) there is countable M∗ ≺ (Hθ,∈, <w) and b ∈ [λ]ω1 ∩M∗ suchthat b ⊇ a, M∗ ⊇ M and M∗ ∩ ω1 = M ∩ ω1. Let CC∗∗ be the statement thatCC∗(λ) holds for all λ ≥ ω2.

A proof structured very much like the above proof that RC implies CC∗ will giveus the following more general fact.

Theorem 7.11 ([22]). Rado’s Conjecture implies CC∗∗.

15More generally, for a cardinal θ > ω1, we let WRP(θ) be the principle saying that for every

stationary set S ⊆ [θ]ℵ0 and every set ω1 ⊆ X ⊆ θ of cardinality ℵ1 there is a subset X ⊆ Y ⊆ θalso of cardinality ℵ1 such that the intersection S ∩ [Y ]ℵ0 is stationary in [Y ]ℵ0 (see [39],[137]).

16More precisely, [91] establishes that WRP(θ) implies θℵ0 = θ for regular cardinals θ ≥ ω2.

12 STEVO TODORCEVIC

So, in particular (see [39], [23]), Rado’s Conjecture implies that ideal NSω1 ofall nonstationary subsets of ω1 is precipitous, i.e. that all its generic ultrapowersare well-founded. It turns out that this can be improved to show that Rado’sConjecture implies that the ideal NSω1 is in fact presaturated, another well-knownstatement with a considerable large cardinal strength. We recall the definition.

Definition 7.12. We say that the ideal NSω1 of non-stationary subsets of ω1 is pre-saturated if for every sequence An (n < ω) of maximal antichains of the quotientalgebra P(ω1)/NSω1 and every stationary S ⊆ ω1 there is stationary T ⊆ S suchthat |A ∈ An : A ∩ T /∈ NSω1| ≤ ℵ1 for all n < ω.

Theorem 7.13 ([36]). Rado’s Conjecture implies that the ideal NSω1 of nonsta-tionary subsets of ω1 is presaturated.

We finish this section with a natural question.

Question 7.14. Does the Strong Chang Conjecture CC∗∗ imply the Singular Car-dinals Hypothesis?

8. An Application of Rado’s Conjecture to Erdos-Hajnal Graphs

To every cardinal θ we associate the Erdos-Hajnal graph

EH(θ) =(ωθ,⊥

),

where we set f ⊥ g iff α < θ : f(α) = g(α) is bounded in θ. Similarly one definesthe other versions of the graph:

EH(λθ) = (λθ,⊥).

Thus, EH(θ) = EH(ωθ). The following fact shows that if the graph EH(θ) is un-countably chromatic than θ must be smaller than the compactness number κG forthe class G of all graphs.

Theorem 8.1 ([26]). If θ is regular then every subgraph of EH(θ) of cardinality< θ is countably chromatic.

Proof. Given a subset X of ωθ of cardinality < θ, find δ < θ such that wheneverf, g ∈ X are such that f ⊥ g then f(γ) 6= g(γ) for all γ ≥ δ. Then f 7→ f(δ) is agood coloring on X showing that (X ,⊥) is countably chromatic.

On the other hand, the following fact shows that EH(θ) is likely to detect any otherincompactness of the countable chromaticity at the level θ.

Theorem 8.2 ([26]). Suppose that a graph G = (θ, EG) on the vertex set θ has theproperty that all of its restrictions Chr(G γ) (γ < θ) are countably chromatic.Then G admits a homomorphism into EH(θ). So in particular, Chr(EH(θ)) ≥Chr(G) for any such graph G = (θ, EG).

Proof. For each γ < θ fix a coloring gγ : γ → ω witnessing Chr(γ,EG ∩ [γ]2) ≤ ℵ0

Define Φ : G→ ωθ by setting

Φ(α)(γ) =

0 if α ≥ γgγ(α) if α < γ

Then α, β ∈ EG ⇒ Φ(α) ⊥ Φ(β), i.e. Φ is a homomorphism. So if G is ofchromatic number ≥ λ then also Chr(EH(θ)) ≥ λ.


Question 8.3 ([26]). For which θ is Chr(EH(θ)) > ℵ0?

Note that Chr(EH(ω)) = 2ℵ0 and Chr(EH(ω1)) ≥ ℵ2, so the first interesting caseof this question is the following.

Question 8.4 ([26]). Is EH(ω2) uncountably chromatic?

Theorem 8.5 ([65]). In some models of set theory Chr(EH(ω2)) can be large. Inparticular, it is possible to have Chr(EH(ω2)) = 2ℵ2 .

We also have the following result which shows that forcing axioms also have someinfluence on the chromatic number of EH(ω2).

Theorem 8.6 ([111]). PFA imlies that Chr(EH(ω2)) ≥ ℵ2.

Sketch of Proof: Using PFA we shall construct a coloring c : [ω2]2 → 2 such that:(1) There is no A ⊆ ω2 of cardinality ℵ2 such that c′′[A]2 = 0.(2) For every α < β < ω2, the set ξ < α : c(ξ, α) = c(ξ, β) = 1 is finite.

Fix an enumeration Xξ (ξ < ω2) of all bounded subsets of ω2. Recursively onα < ω2, we shall choose sets Aα ⊆ α such that

(3) Aα ∩Aβ ∈ Fin whenever α 6= β,(4) Aβ ∩ Xξ 6= ∅ for all ξ < β with the property that Xξ \ (

⋃α∈D Aα) is

uncountable for all countable D ⊆ β.If for some β < ω2, the sequence Aα (α < β) is constructed, choose Bβ ⊆ βsatisfying (4) in place of Aβ and the weaker form of (3) to the effect that Aα ∩Bβis countable when α < β. Now define the standard ccc poset to thin out Bβ to asubset Aβ satisfying (3) and (4). When finished, set

c(α, β) = 1 whenever α ∈ Aβ .

It is clear that c has the desired properties. Consider the graph

G = (ω2, c−1(1)).

Then G has no independent set of size ℵ2, so in particular, its chromatic numberis equal to ℵ2. On the other hand, the property (2) of the coloring c translatesinto the fact that the poset P of all finite G-independent subsets of ω2 satisfies thecountable chain condition. So applying MAℵ1 we conclude that

Chr(G α) ≤ ℵ0 for all α < ω2.

By Theorem 8.2, we conclude that G ≤ EH(ω2), as required.

Problem 8.7. Does PFA imply that Chr(EH(ω2)) > ℵ2?

Let us now give some upper bounds on Chr(EH(θ)). For this we need the follow-ing standard notion.

Definition 8.8 ([51]). A poset P is said to satisfy σ-finite chain condition if itadmits a countable decomposition

P =⋃n<ω

Pn

with the property that no Pn contains an infinite set of pairwise incompatibleelements.

14 STEVO TODORCEVIC

Proposition 8.9. If θ supports a proper uniform σ-complete ideal I with the prop-erty that the corresponding quotient algebra P(θ)/I satisfies the σ-finite chain con-dition, then Chr(EH(θ)) ≤ ℵ0.

Corollary 8.10. Chr(EH(θ)) ≤ ℵ0 for every real-valued measurable cardinal θ.

Proposition 8.11. If U is a uniform ultrafilter on θ, then

Chr(EH(θ)) ≤∣∣ωθ/U∣∣ .

The following result shows that this hypothesis can sometimes be achieved withthe minimal possible value for the ultrapower.

Theorem 8.12 ([38]). It is consistent relative to the existence of some large car-dinals that |ωω2/U| = ℵ1 for some uniform ultrafilter U on ω2.

Corollary 8.13 ([38]). It is consistent that Chr(EH(ω2)) ≤ ℵ1.

Proposition 8.14. Shift(EH(cθ)) ≤ EH(ωθ) = EH(θ).

Proof. Let qn : n < ω be an enumeration of the rationals. Define ∆ : R2 → ω bysetting

∆(x, y) =

0 if x = y

n with n minimal such that qn ∈ (x, y) or qn ∈ (y, x).

DefineΦ : (f, g) ∈ (Rθ)2 : f ⊥ g and f <lex g → Z

θ

by lettingΦ(f, g)(ξ) = ∆ (f(ξ), g(ξ)) · sgn (f(ξ)− g(ξ))

Note that Φ(f, g) ⊥ Φ(f ′, g′) whenever f <lex g = f ′ <lex g′ and f ⊥ g andf ′ ⊥ g′. It follows that Φ is a homomorphism from Shift(EH(Rθ)) into EH(θ), asrequired.

Corollary 8.15. Chr(EH(cθ)) > c implies Chr(EH(θ)) > ℵ0

Proof. This follows from Proposition 8.14 and Theorem 2.4.

Theorem 8.16 ([119]). Chr(EH(ω2)) > ℵ0.

Proof. Note that Kc+ ≤ EH(cω2) when c ≤ ℵ2. So by Corollary 8.15, we get thedesired conclusion Chr(EH(ω2)) > ℵ0 in case c ≤ ℵ2. So it remains to reach thesame conclusion when c > ℵ2. Fix

e : [ω2]2 → ω1

such that e(α, γ) 6= e(β, γ) whenever α < β < γ < ω2.Recall the following two properties of the family Fe of intervals of the linearly

ordered set Le constructed above:(1) Fe =

⋃ξ<ω2

Fξe such that Fξe ⊆ Fηe whenever ξ < η, and such that for everyξ < ω2 the family Fξe can be decomposed into countably many familiesconsisting of pairwise disjoint intervals.

(2) Since R * L[e], the family Fe itself cannot be decomposed into countablymany subfamilies consisting of pairwise disjoint intervals.


Let (Fe, Int) be the intersection graph of Fe: (I, J) ∈ Int if and only if I ∩ J 6= ∅.For every ξ < ω2, we fix a map gξ : Fξe → ω such that I ∩ J 6= ∅ ⇒ gξ(I) 6= gξ(J).Define Φ : Fe → ωω2 by setting

Φ(I)(ξ) =

1 if I /∈ Fξegξ(I) if I ∈ Fξe .

Claim 1. Φ : (Fe, Int) → (EH(ω2),⊥) is a homomorphism, or in other words,I ∩ J 6= ∅ implies Φ(I) ⊥ Φ(J).

It follows that Chr(EH(ω2)) ≥ Chr(Fe, Int). Since by (2), Chr(Fe, Int) > ℵ0, theproof of Theorem 8.16 is finished.

Problem 8.17. Investigate the chromatic number of EH(θ) =(ωθ,⊥

)for other

values of θ such as for example θ = ω3, ω4, ω5, . . . , or θ = ωω+1.

Theorem 8.18 ([116]). For every infinite cardinal θ, the inequality Chr(EH(θ)) ≤ℵ0 implies θ → (ω1)<ωω1,ω

. 17 In particular, it there is an infinite cardinal θ such thatChr(EH(θ)) ≤ ℵ0 then 0] exists.

Sketch of Proof: The proof will follow closely the proof of Theorem 7.1 above. ForA ∈ [ω1 ∪ ωθ1 ]ℵ0 , set

DA = α < θ : (∀f ∈ A)f(α) ∈ A.Let

Sθ = A ∈ [ω1 ∪ ωθ1 ]ℵ0 : A ∩ ω1 ∈ ω1 and sup(DA) < θ.As before, it suffices to show that Chr(EH(θ)) ≤ ℵ0 implies that Sθ is not

stationary. (Note that our assumption Chr(EH(θ)) ≤ ℵ0 implies that in particularcf(θ) > ω1. )

Let T be the collection of all sequences t = (tξ : ξ ≤ `(t)) such that:(a) `(t) < ω1 and tξ ∈ Sθ for all ξ ≤ `(t),(b) tξ ⊆ tη and tξ ∩ ω1 < tη ∩ ω1 for all ξ < η ≤ `(t),(c) tλ =

⋃ξ<λ tξ for all limit λ ≤ `(t).

We consider T as a tree ordered by end-extension. For δ < θ, let

T δ = t ∈ T : (∀ξ ≤ `(t)) Dtξ ⊆ δ.

Claim 1. For every δ < θ, the subtree T δ can be covered by countably manyantichains.

Proof. Fix δ < θ. By the Pressing Down Lemma for trees, it suffices to define aregressive mapping

H : T δ → T δ

such that every of its pre-images H−1(s) s ∈ T δ allows a decomposition into count-ably many antichains. For t a successor node of T δ, let H(t) be its immediatepredecessor. Consider now t ∈ T δ of limit height `(t). Since δ 6∈ Dtξ for ξ = `(T )there is a function

ft ∈ t`(t) ∩ ωθ1 such that ft(δ) ≥ t`(t) ∩ ω1.

Let H(t) be the minimal initial segment s of t such that ft ∈ s`(s).

17Recall that θ → (ω1)<ωω1,ωmeans that for every mapping f : θ<ω → ω1 there exists uncount-

able X ⊆ θ such that |f ′′X<ω | ≤ ℵ0.

16 STEVO TODORCEVIC

We need to show that for every s ∈ T δ, the pre-image H−1(s) can be decomposedinto countably many antichains. Fix an s ∈ T δ. For f ∈ t`(s) ∩ ωθ1 , set

Wf = t ∈ H−1(s) : ft = f.Since H−1(s) =

⋃f∈s`(s) Wf is a countable union, it suffices to show that each set

Wf can be covered by countably many antichains. To see this first observe that

f(δ) ≥ t`(t) ∩ ω1 for all t ∈Wf .

According to the condition (b) from the definition of T δ, this in particular meansthat Wf contains no chain of order type > f(δ). So Wf as a subtree of T δ hasonly countably many levels, so in particular can be covered by countably manyantichains.

For δ < θ, fix Gδ : T δ → ω such that G−1δ (n) is an antichain for all n < ω. Define

Φ : T → ωθ, by

Φ(t)(δ) =

0 if t /∈ T δ

Gδ(t) if t ∈ T δ

Then Φ is a homomorphism from the comparability graph of T into EH(θ). Itfollows, in particular, that the comparability graph of T is countably chromatic,or equivalently, that the whole tree T can be decomposed into countably manyantichains. So the proof is finished, once we show the following.

Claim 2. If Sθ is stationary, the corresponding tree T cannot be decomposed intocountably many antichains.

Proof. Consider a decomposition T =⋃n<ω Xn. Assuming that Sθ is stationary we

will show that one of the Xn’s is not an antichain. Choose a countable elementarysubmodel M of H(2θ)+ containing all these objects such that M ∩ (ω1 ∪ ωθ1) ∈ Sθ.Then it is possible to build an increasing sequence tn : n < ω in T ∩M with anupper bound t ∈ T such that:

(1) t`(t) = M ∩ (ω1 ∪ ωθ1),(2) for every n < ω, either tn ∈ Xn or else no extension of tn belongs to Xn.

Pick an n < ω such that t ∈ Xn. Then tn must also belong to Xn. So, Xn containsto distinct comparable nodes t and tn.

Corollary 8.19. In Godel’s constructible universe L, Chr(EH(θ)) > ℵ0 for allinfinite cardinals θ.

Problem 8.20. Evaluate Chr(EH(θ)) in the Godel constructible universe L.

Corollary 8.21. The weak compactness of an infinite cardinal θ is not sufficientfor the conclusion Chr(EH(θ)) ≤ ℵ0.

18

9. The Countable Chain Condition on Graphs and Hypergraphs

Recall that a hypergraph is a structure of the form H = (X,E) where X is a setand E a collection of nonempty finite subsets of X. When every element of E hassome fixed size n, we call H = (X,E) an n-hypergraph. Thus a graph is nothingmore than a 2-hypergraph. Elements of E are called hyperedges. A subset D of Xis H-discrete if it includes no hyperedge.

18Recall that measurability of θ is sufficient for this conclusion.


Definition 9.1. The chromatic number, Chr(H), of a hypergraph is the minimalcardinality of a partition of X into H-discrete subsets of X. Note that this agreeswith the notion of chromatic number for graphs.

It is well known that forbidding complete subgraphs (or even K3) is not a suffi-cient condition for an upper bound on the chromatic number of a given graph.It turns out however that the following natural strengthening of the conditionKℵ1 6≤ H has some facility for giving us such an upper bound.

Definition 9.2. A hypergraph H = (X,E) satisfies the countable chain conditionif for every uncountable family F of finite H-discrete subsets of X there exist p 6= qin F whose union p ∪ q is also H-discrete.

So in this context the following statement falls naturally into the category of graph-theoretic dichotomy results of our interest here.

Definition 9.3 ([112]). [The statement Σ] For every hypergraph H = (X,E) on aset X of cardinality at most continuum, either H fails to satisfy the countable chaincondition, or else Chr(H) ≤ ℵ1. If we restrict Σ to n-hypergraphs, the correspondingweaker (though in some sense more interesting) statement is denoted by Σn. Therestrictions to the corresponding classes of Borel hypegraphs19 are denoted by ΣBor ΣnB , respectively.

Question 9.4. Is any of the statements Σ2, Σ3, · · · , Σ equivalent to CH?

It has been proved in [112] that ΣB is consistent with the continuum large so thereis no way to capture the full strength of CH by using just Borel structures 20.However the following sample results show that Σ and ΣB do capture basically allthe standard consequences of CH.

Theorem 9.5 ([112]). Σ implies that every function f : ω2 × ω2 → ω must beconstant on some product of two infinite subsets of ω2.

Theorem 9.6 ([112]). Σ implies that if some inner model satisfies MAω2 then 0]

exists.

Theorem 9.7 ([112]). Σ implies that every free ultrafilter U on N has π-characterℵ1, i.e., there is a family F of ℵ1 subsets of N such that every element of U includesone of F .

Sketch of Proof. Apply Σ to the hypergraph H = (U , E), where E is the collectionof all finite subsets X of U for which we can find an integer k such that

|(⋂X ) ∩ 0, 1, ..., k − 1| < |∆(A,B) : A,B ∈ X , A 6= B|.

Note that while the vertex-set U of this hypergraph is not Borel its hyperedgerelation is. So this application of Σ allows large continuum.

Theorem 9.8 ([112]). ΣB implies each of the following standard consequences ofthe Continuum Hypothesis:

19A hypergraph H = (X,E) is Borel if X is a Polish space and E is a Borel collection of finitesubsets of X.

20It is shown in [112] that large continuum is consistent with Σ applied to any hypergraphH = (X,E) whith X a set of reals and E = E∗ ∩ [X]<ω where E∗ is a Borel set of finite sets ofreals. Thus, the vertex set X need not be Borel.

18 STEVO TODORCEVIC

(a) There is a family F of ℵ1 nowhere dense sets of reals such that everynowhere dense set of reals is included in a member from F .

(b) There is a family F of ℵ1 measure zero sets of reals such that every measurezero set of reals is included in a member from F .

Sketch of Proof. Since the ideal of meager subsets of R is Tukey-reducible to theideal of measure zero subsets of R (see [42] and Section 24.1 below) the statement(a) is a consequence of the statement (b). To see that ΣB implies (b), consider thehypergraph H = (A, E), where A is the collection of all compact subsets of R ofmeasure > 1/2 and where E is the family of all finite subsets K of A such that

⋂K

has measure ≤ 1/2.

The more interesting finite-dimensional Σn have similar though weaker consequences.

Theorem 9.9 ([112]). Σ3 implies cf(c) = ℵ1.

Sketch of Proof. To a given function f : NN → NN we associate the triple system

H = (NN, Ef ) by letting a, b, c ∈ Ef if and only if either f a, b, c is notone-to-one, or

|∆(a, b),∆(a, c),∆(b, c)| > |∆(f(a), f(b)),∆(f(a), f(c)),∆(f(b), f(c))|.21

Then Hf satisfies the countable chain condition and the function f is continuous onany subset X of NN which includes no triple from Ef . Now apply Σ2 to the triplesystem Hf associated to a one-to-one function f : NN → N

N that is not continuouson any set of size continuum.

Theorem 9.10 ([112]). Σ3B implies d = ℵ1.

Sketch of Proof. Apply Σ3B to the Borel triple systemH = (NN, Ed) where Ed is the

collection of all triples a, b, c ⊆ NN such that |∆(a, b),∆(a, c),∆(b, c)| = 1.

Theorem 9.11 ([112]). Σ2 implies b = ℵ1.

Sketch of Proof. Choose a subset X of NN consisting of increasing functions fromN in N such that X is well-ordered and unbounded relative to the ordering ofeventual dominance of NN. Apply Σ2 to the graph H = (X,Eb), where Eb is thecollection of all 2-element subsets of X such that either (∀n) a(n) ≤ b(n) or else(∀n) b(n) ≤ a(n).22

Part 2. Topologically Presented Graphs

In this part we look at compactness of chromatic numbers of topologically pre-sented graphs, again with the aim of finding classes with small compactness numbersand interesting consequences. Thus, we shall be interested in graphs of the form

G = (X,E),

where X is a topological space and E a symmetric irreflexive subset of X2 with someprescribed topological complexity such as, for example, open, closed, Fσ, etc. Wehave already seen above that spaces X with large discrete subspaces will support

21Note that no matter what the function f is, the triple system Hf can be represented in theform (X,E) where X is some set of reals and where E is the restriction to X of some Borel triplesystem on R. So this application of Σ3 also alows the continuum to be large.

22Note that while X is not Borel, the relation Eb is Borel. So the form of Σ2 used here alsoallows the continuum to be large.


graphs of small complexity but large compactness number. In fact, at the currentstage of our knowledge, we can say something only when X is a separable metricspace. So from now on, the vertex set X of a topologically given graph G = (X,E)will be assumed to be at least separable metric. The following well known fact23

shows that if we wish to keep the vertex set to be an arbitrary separable metricspace we must add severe restrictions on the complexity of the edge-relation.

Proposition 9.12 ([40]; 21F). For every symmetric irreflexive relation E ⊆ p2

there is a sequence aξ : ξ < p ⊆ [ω]ω such that (ξ, η) ∈ E iff aξ ∩ aη ∈ Fin.

In other words, an arbitrary graph on a vertex set of cardinality at most p isisomorphic to a graph whose vertex set is a separable metric space and whose edgerelation if Fσ. Thus, we are restricted to considering the following two compactnessnumbers.

Definition 9.13. The open graph compactness number, κO, is the minimal numberκ with the property that every open graph G = (X,E) on a separable metric spaceis countably chromatic if all of its subgraphs of cardinality < κ are countablychromatic. The closed graph compactness number, κC, is the minimal number κwith the property that every closed graph G = (X,E) on a separable metric spaceis countably chromatic if all of its subgraphs of cardinality < κ are countablychromatic.

Note that we always have the inequalities c+ ≥ κO and c+ ≥ κC. So in this contextthe cardinals have their nontrivial realizations only when they are smaller or equalto the continuum.

Problem 9.14. Are κO and κC regular cardinals whenever nontrivially realized?

10. The Open Graph Compactness Number and the Bounding number

Recall the following important cardinal characteristic of the continuum

b = min|F| : F ⊆ N

N,(∀g ∈ NN

)(∃f ∈ F) f ≮∗ g

.

Note that this is a regular cardinal and note the following immediate consequenceof the definition of this cardinal.

Proposition 10.1. There is always a <∗-well-ordered <∗-unbounded chain of ordertype b in (NN, <∗).

The purpose of this section is to prove the following.

Theorem 10.2 ([108]). b ≤ κO if κO is a regular cardinal and b ≤ κ+O if κO is a

singular cardinal.

Proof. Assume the contrary, that b > θ, where θ be the minimal regular cardinal≥ κO. By Proposition 10.1, we can choose a <∗-increasing <∗-unbounded sequenceA = aξ : ξ < θ in (NN, <∗) consisting of strictly increasing mappings from N intoN. Since A is bounded in (NN, <∗) we can choose a maximal <∗-chain B consistingof upper bounds of A. Note that if b is an upper bound of A then so is b − 124

and so by its maximality the set B has no <∗-minimal element. Therefore, by our

23where p is the minimal cardinality of a family F ⊆ [ω]ω that is closed under finite intersection

such that there is no infinite b ⊆ N such that b \ a ∈ Fin for all a ∈ F .24defined by (b− 1)(n) = max(0, b(n)− 1).

20 STEVO TODORCEVIC

assumption b > |A|, the chain (B,<∗) must have uncountable coinitiality. Let B∗

be the set of all finite changes of elements of B, and let

X = (a, b) ∈ A×B∗ : (∀n) a(n) ≤ b(n).Then X is a separable metric space with the topology induced from N

N × NN.

Consider the following open edge relation on X,

E = ((a, b), (a′, b′)) ∈ X2 : (∃n)a(n) > b′(n) or (∃n)a′(n) > b(n).This gives us the open graph G = (X,E) that will be subject to the definition ofκO once we establish the following.

Claim 1. Every subgraph G Y of cardinality < κO is countably chromatic.

Proof. Choose c in A such that a <∗ c for every a appearing as a first coordinateof an element of Y. Let ck : k < ω lists all finite changes of c. For k < ω, let

Yk = (a, b) ∈ Y : (∀n) a(n) ≤ ck(n) ≤ b(n)Then Y =

⋃k<ω Yk is a coloring witnessing Chr(G Y ) ≤ ℵ0.

By the definition of κO, it follows that Chr(G) ≤ ℵ0. So let X =⋃k<ωXk be

a good coloring witnessing this. Since B has uncountable coinitiality, there mustbe k such that the first projection Ak of Xk is cofinal in (A,<∗) while the secondprojection Bk of Xk is coinitial in (B∗, <∗). Define c ∈ NN, by

c(n) = minb(n) : b ∈ Bk.Then A <∗ c <∗ B, and so c∪B contradicts the maximality of B. This completesthe proof

11. A conjecture about open graphs on separable metric spaces

The purpose of this section is to expose the following conjecture that is quiteanalogous to Galvin’s and Rado’s conjectures considered above in Part I.

Definition 11.1 ([108]). [Todorcevic’s Conjecture25] κO = ℵ2.26

.Clearly, CH implies TC for the trivial reason that in this case we have the equalityκO = c+. So since we are interested in the nontrivial realizations of κO we haveto look for appearances of TC in the context of the negation of CH. The followingresult establishes one such case.

Theorem 11.2 ([108]). The Proper Forcing Axiom implies that κO = ℵ2.

Remark 11.3. More precisely, we shall show that given an open graph G = (X,R)on a separable metric space X such that Chr(G) > ℵ0 there is a proper poset whichforces Kℵ1 ≤ G.

[Sketch of Proof of Theorem 11.2] Let P be the collection of all pairs p = 〈Hp,Np〉with the following properties:

25The Todorcevic Conjecture, TC in short, shows up implicitly first in [108] as the main partof an axiom about open graphs introduced there (see also Definition 11.6 below). The name wassuggested by the Editor of this Bulletin.

26In other words, an open graph on a separable metric space is countably chromatic if andonly if all of its subgraphs of cardinality at most ℵ1 are countably chromatic.


(1) Hp is a finite clique of G, i.e., H is a finite subset of X such that for everyx 6= y ∈ Hp the pair x, y is an edge of G,

(2) Np is a finite ∈-chain of countable elementary submodels N of (Hc+ ,∈),(3) Np separates Hp, that is,

(∀x 6= y ∈ Hp) (∃N ∈ Np) |N ∩ x, y| = 1,

(4) (∀N ∈ Np)(∀x ∈ Hp \N)(∀Y ∈ N)[x ∈ Y → Y 2 ∩R 6= ∅].and we set p ≤ q if and only if

(5) Hp ⊇ Hq and Np ⊇ Nq.Theorem 11.2 follows easily from the following Claim.

Claim 1. P is proper.

[Sketch of the proof of Claim 1] Let p ∈M ≺(H(2c)+ ,∈

). We prove that

q = (Hp,Np ∪ M ∩Hc+)

is an (M,P)-generic condition. That is, we need to show

(∀D ∈M such that D ⊆ P is dense open) (∀r ≤ q) (∃r ∈ D ∩M) r ∼ r.27

We may assume r ∈ D. Let n = |Hr \M | . Let

xr =⟨xr0, x

r1, . . . , x

rn−1

⟩,

whereHr \M =

xr0, x

r1, . . . , x

rn−1

,

with indexing chosen according to the ∈-ordering of Nr:

(∀i < n− 1) (∃N ∈ Nr) [xi ∈ N & xi+1 /∈ N ] .

LetF = x ∈ Xn : ∃s ≤ r ∩M [s ∈ D & xs = x] .

Then F ∈M and the family F is large in the following precise sense.

Subclaim 1.1. The family F of n-tuples is Namba relative to the co-ideal

Q = Y ⊆ X : Chr (G Y ) > ℵ0 ,

or in other words the following formula is true

Qx0Qx1 . . . Qxn−1 〈x0, x1, . . . , xn−1〉 ∈ F .28

Note that Namba families of n-tuples of elements of X form a nontrivial σ-completecoideal of subsets of Xn.

Subclaim 1.2. For almost all x ∈ F in the sense of Namba, there is y ∈ F suchthat

∀i < n [(xi, yi) ∈ R] .

27Here, ∼ denotes the compatibility relation of P.28Recall that for a formula ϕ(v) with one free variable v ranging over the set X, the formula

Qxϕ(v) is interpreted as saying that x ∈ X : ϕ(x) ∈ Q.

22 STEVO TODORCEVIC

Proof. Induction on n. Let

H0 =x ∈ Xn−1 : Chr (G Yx) > ℵ0 for Yx = a ∈ X : x_a ∈ F

.

Then from the assumption that F is a Namba subset of Xn, we conclude that H0

is a Namba subset of Xn−1. Fix a countable base B ∈ M for X. Since for x ∈ H0

we have that in particular Y 2x ∩R 6= ∅, and since R is open, we can fix Ux, Vx ∈ B

such that(1) Ux ∩ Vx = ∅,(2) Ux ∩ Yx 6= ∅ 6= Vx ∩ Yx,(3) Ux × Vx ⊆ R.

Find Namba H1 ⊆ H0 and U, V ∈ B such that

(∀x ∈ H1) [Ux = U & Vx = V ] .

By the inductive hypothesis there exist x, y ∈ H1 such that

(∀i < n− 1) (xi, yi) ∈ R.Find a ∈ U such that x_a ∈ F0. Find b ∈ V such that y_b ∈ F0. Then x_a andy_b are as required.

Note that this finishes the proof of Claim 1 as well as the proof of Theorem 11.2.As it will be clear from an example of a closed graph given below (see Example

19.1), we have the following result which shows that the nontrivial realization ofthe closed graph compactness number is incompatible with PFA.

Theorem 11.4. PFA implies that the closed graph compactness number κC hasonly the trivial realization, or more precisely, that κC = c+.

It turns out that TC is a compactness principle with strong influence on struc-tures that are related in some way to the set of real numbers. For example, notethe following consequence of Theorem 10.2.

Theorem 11.5 ([108]). TC implies b ≤ ℵ2.

It turns out that TC does not imply c ≤ ℵ2 since adding any number of Cohenreals to a model of CH produces a model of TC while the continuum is arbitrarilylarge (see [29]).

As remarked above (Remark 11.3), the proof of Theorem 11.2 gives the followingstronger form of the principle TC proven to be of considerable and of independentinterest.

Definition 11.6 ([108]). [Todorcevic’s Axiom] Let TA denote the statement thatfor every open graph G = (X,E) on a separable metric space X either Chr(G) ≤ ℵ0,or Kℵ1 ≤ G.

Thus, TA is a strengthening of TC in the sense that TA = TC + TAℵ1 , whereTAℵ1 denotes the restriction of TA to separable metric spaces of cardinality atmost ℵ1. The name was again suggested by the Editor of this Bulletin (see, also,[35]). Originally in [108] we have used the name Open Coloring Axiom for TAbut the name has already been used in [1] for a rather different axiom which, inour terminology here, applies to graphs G = (X,E) on separable metric spacesX of size ℵ1 with edge relations E clopen rather than open and asserts that thevertex set X can be covered by countably many sets that are either G-complete orG-discrete. It should also be noted that the paper [1] does have an axiom, denoted


there by SOCA1 (see [1], p.136), which is in fact stronger than our TAℵ1 but whichunfortunately fails if one does not restrict sets to be of cardinality at most ℵ1 (see,for example, Example 19.1 below).

We have already noted that the proof of Theorem 11.2 establishes also the fol-lowing fact which we now state for the record.

Theorem 11.7 ([108]). The Proper Forcing Axiom implies TA.

12. The Open Graph Axiom, the Bounding Number, and the Continuum

Theorem 12.1 ([108]29). TA implies b = ℵ2.

Question 12.2. Does TA imply c = ℵ2?

We give some details from the proof of Theorem 12.1 as they are of independentinterest. There are two ingredients in the proof of the equality b = ℵ2 from TA:

(1) Gaps in quotient structures P(N)/Fin, NN/Fin, . . . .(2) Oscillation theory of NN and of P(N).

We start by first describing the basic ideas behind the oscillation theory as de-veloped in [107] (see also [108], [111], and [76].) For x, y ∈ N

↑≤N (the set of allnondecreasing finite or infinite sequences of integers), let

D(x, y) = n : x(n) 6= y(n) .

Let E(x, y) be an equivalence relation on D(x, y) defined by

m E(x, y) niff

∀k ∈ [m,n] (x(m) < y(m) ⇐⇒ x(k) < y(k) ⇐⇒ x(n) < y(n)).

Letosc(x, y) = |D(x, y)/ E(x, y)| .

Theorem 12.3 ([107]). Suppose X is a totally ordered and unbounded subset of(N↑N, <∗). Then

(∀k < ω) (∃x, y ∈ X) [osc(x, y) = k] .

Definition 12.4. An equivalence class e ∈ D(x, y)/ E(x, y) is said to be relativelylarge if

|e| > |x(n)− y(n)| for n = min(e).

Remark 12.5. Note that if x <∗ y, one of the classes of D(x, y)/ E(x, y) is infinite,so in particular there is a relatively large class. In fact the proof of the previousresult gives x, y with osc(x, y) = k such that all classes of D(x, y)/ E(x, y) arerelatively small except the last one which is infinite. So if we let

osc∗(x, y) = osc (x n, y n) ,

where n is the minimum of the first relatively large class of D(x, y)/ E(x, y), weget the following reformulation of Theorem 12.3.

29This result appears in [108] as two separate statements (Theorems 8.5 and 3.7, respectively)TA⇒ A and A⇒ b = ℵ2, where A is a topological basis theorem.

24 STEVO TODORCEVIC

Theorem 12.6. Suppose X is a totally ordered and unbounded subset of (N↑N, <∗).Then

(∀k < ω) (∃x, y ∈ X) [osc∗(x, y) = k] .

Corollary 12.7. TA implies b > ℵ1.

Proof. Choose a totally ordered and unbounded subsetX of (N↑N, <∗) of cardinalityb. Consider the open graph G = (X,E), where we put x, y ∈ E if osc(x, y) > 0(or if osc∗(x, y) > 0).

Remark 12.8. It follows that under TA there is a sequence xξ : ξ < ω2 ⊆ N↑N

such that xξ <∗ xη whenever ξ < η < ω2.

Question 12.9. Given such an increasing sequence (xξ : ξ < ω2), and assuming ifnecessary TA, when does the inner model L [xξ : ξ < ω2] contain all the reals?

Theorem 12.10 ([110]30). Assume TA and fix an <∗-unbounded <∗-chain

xξ : ξ < ω2 ⊆ N↑N.

Suppose further that M ⊇ L [xξ : ξ < ω2] is an inner model of a sufficiently largefragment of ZFC with the property that every uncountable subset of ω1 contains aninfinite subset belonging to M . Then M contains all the reals.

Sketch of Proof. Let us say that an infinite set X ⊆ N↑N codes a real r ∈ 2N if 31

(∀x 6= y ∈ X) (∃k ≥ ∆(x, y)(= minD(x, y))) osc∗(x, y) =k∑i=0

r(i)2i.

Let X = xξ : ξ < ω2, fix a real r ∈ 2N \ Fin, and assume TA.Define Er ⊆ X2 \∆ by

(x, y) ∈ Er ⇐⇒ (∃k ≥ ∆(x, y)) osc∗(x, y) =k∑i=0

r(i)2i.

Clearly Er is open. By the basic oscillation result, Theorem 12.6, we have Y 2∩Er 6=∅ for all unbounded Y ⊆ X, so in particular

Chr(X,Er) > ℵ0.

Applying TA, we get an uncountable Y ⊆ X coding r. Shrinking Y we may assumethat the closure Y is a compact subset of NN. By our assumption about M there isinfinite Y0 ⊆ Y such that Y0 ∈M . By compactness of the closure of of Y, we havethat sup∆(x, y) : x, y ∈ Y0, x 6= y =∞, It follows that r ∈M .

Corollary 12.11. TA implies that the continuum is bounded by

min|F| : F ⊆ [ω2]ℵ0 and (∀X ∈ [ω2]ℵ1)(∃D ∈ F) D ⊆ X.Definition 12.12. For Y ⊆ N

↑N, let

R(Y ) =r ∈ 2N : r is coded by some uncountable Z ⊆ Y

.

Problem 12.13. Assuming TA, is there a totally <∗-ordered <∗-unbounded chainX = xξ : ξ < ω2 ⊆ N

N such that |R(X)| ≤ ℵ2?

30This result was announced in print in [16], p. 168 in a slightly weaker formulation (that PFA

fails in the Sacks forcing extension), but this is what the proof gives.31Our original definition was slightly different (see [11], [119]), the current one is taken from

[76].


13. The Open Graph Axiom and Gaps in P(ω)/Fin

Definition 13.1. For a, b ⊆ ω, set

a ⊥ b if and only if a ∩ b ∈ Fin.

For A,B ⊆ P(ω), set

A ⊥ B if and only if (∀a ∈ A) (∀b ∈ B) a ⊥ b.For A ⊆ P(ω), set

A⊥ = b : (∀a ∈ A) b ∩ a ∈ Fin.We say that c ⊆ ω separates A and B if and only if

(∀a ∈ A) (∀b ∈ B) [a ⊆∗ c & c ⊥ b] .We say that C ⊆ P(ω) separates A and B if and only if

(∀a ∈ A) (∀b ∈ B) (∃c ∈ C) [a ⊆∗ c & c ⊥ b] .

Example 13.2 ([48],[49]). There exist two orthogonal ω1-chains A = aξ : ξ < ω1and B = bξ : ξ < ω1 in the quotient algebra P(ω)/Fin which cannot be separated.This is what is usually called Hausdorff’s (ω1, ω1)-gap in P(ω)/Fin.

Definition 13.3. An indexed family (ai, bi) : i ∈ I ⊆ P(ω) × P(ω) is said toform a biorthogonal gap if

(ai ∩ bj) ∪ (aj ∩ bi) = ∅ iff i = j

Remark 13.4. While this is inspired by the Hausdorff condition on gaps, notethat this notion corresponds to countable separation rather than separation. Moreprecisely, note that a biorthogonal gap indexed by an uncountable set I cannot beseparated by a countable C ⊆ P(ω).

It turns out that TA has a considerable influence on the gap structure of thequotient algebra P(ω)/Fin as the following fact shows.

Theorem 13.5 ([108]). Assume TA. For every pair A and B of downwards closedorthogonal families of subsets of ω, we have that either

(1) some countable C ⊆ P(ω) separates A and B, or(2) there is an uncountable biorthogonal gap (ai, bi) : i ∈ I ⊆ A× B.

Proof. LetX = (a, b) ∈ A× B : a ∩ b = ∅

and letR = ((a, b), (a′, b′)) ∈ X 2 : (a ∩ b′) ∪ (b ∩ a′) 6= ∅.

Then G = (X , R) is an open graph, so applying TA, we have to consider thefollowing two alternatives:

(1) If Chr(G) ≤ ℵ0 then there is a decomposition X =⋃n<ω Xn such that

(Xn)2 ∩ R = ∅ for all n < ω. For n < ω, let cn =⋃a : (∃b) (a, b) ∈ Xn .

Then the family C = cn : n < ω separates A and B.(2) If K ⊆ A⊗ B is a clique of G, i.e. if K2 \∆ ⊆ R then any 1-1 enumerationK = (ai, bi) : i ∈ I forms a biorthogonal gap.

Corollary 13.6. Assuming TA, the orthogonal A⊥ of any chain A of P(ω)/Finof cofinality > ω1 is countably generated.

26 STEVO TODORCEVIC

Remark 13.7. This result is sharp since under p > ω1, an assumption compatiblewith TA, the orthogonal of a chainA of P(ω)/Fin of cofinality ≤ ω1 is not countablygenerated.Corollary 13.8 ([108]). TA implies b ≤ ℵ2.

Proof. If b > ℵ2 then there is an ω2-chain in P(ω)/Fin whose orthogonal is notcountably generated. In fact this is true about any ω2-chain in P(ω)/Fin with noleast upper bound.

14. The Open Graph Axiom and the Homomorphism RepresentationProblem for Quotient Algebras of the Form P(N)/I

This problem is about solving the following diagram

P(N) Φ∗−−−−→ P(N)

πI

y πJ

yP(N)/I Φ−−−−→ P(N)/J ,

where I,J are analytic ideals on N and πI , πJ are the corresponding quotient maps,Φ is a homomorphism we wish to study and Φ∗ is its lifting. In other words, this is arepresentation theory of homomorphisms Φ : P(N)/I → P(N)/J between quotientalgebras over analytic ideals on N in terms of their liftings Φ∗ : P(N) → P(N).Note that Φ∗ may not be a homomorphism itself but when it is one gets a goodsolution to the lifting problem. Particularly desirable are the completely additiveliftings Φ∗, the maps Φ∗ : P(N) → P(N) that preserve even infinitary Booleanoperations. Note that every such Φ∗ has the form Φ∗(x) = h−1(x) for some maph : N→ N witnessing the Rudin-Keisler reduction J ≤RK I.32 Note also that suchmaps Φ∗ are continuous suggesting us to consider also topological requirements onthe liftings. This has turned out to be the right approach as first shown in the caseof I = J = Fin by the ground-breaking work [90].

Theorem 14.1 ( [61], [90], [132]). If a homomorphism Φ : P(N)/Fin→ P(N)/Finhas a Baire-measurable (or equivalently continuous) lifting then it also has a com-pletely additive lifting, so its kernel is an ideal generated over Fin by a single subsetof N.

That this remains true for arbitrary homomorphism Φ : P(N)/Fin→ P(N)/Finin some model of set was also first shown in [90] but the introduction of TA in thisarea was made in [133] and [62] allowing us to have the following result.

Theorem 14.2 ([62],[122],[133]). Assume TA. If I is a non-atomic analytic idealon N then P(N)/I is not isomorphic to a subalgebra of P(N)/Fin.

The ground-breaking work for the general theory was made in [62] building ona series of previous papers [59],[60],[61].

Theorem 14.3 ([62]). TA implies P(N)/Z0 P(N)/Zlog.33

This suggested two directions of study in this area that are based on two differenttheories:

32In other words, for an arbitrary set Y ⊆ N, we have that Y ∈ J if and only if h−1(Y ) ∈ I.

33Here, Z0 = X ⊆ N : limn→∞|A∩n|n

= 0, and Zlog = X ⊆ N : limn→∞

∑k∈A∩n

1k+1

ln n= 0.


ZFC ZFC+MA+TAhomomorphisms with continuous liftings arbitrary homomorphisms

In fact, there is a close connection between these two directions which is supportedby the fact that TA has the power of reducing problems about arbitrary homomor-phisms to problems about homomorphisms with Baire-measurable liftings. Thestudy of the special case of analytic P-ideals was first suggested in [122] inspiredby the following two results.

Theorem 14.4 ([122]). Suppose Φ : P(N)/I → P(N)/J is an isomorphic embed-ding with a continuous lifting. If J is an analytic P-ideal then so is I.

Theorem 14.5 ([122]). Assume TA and that I and J are analytic ideals such thatP(N)/I is embeddable into P(N)/J . If J is a P-ideal then so is I.

This turned out to be a fruitful approach as shown in another ground-breakingwork [31] where it is proved that in the case of analytic P-ideals I and J , thelifting problem from [122] is equivalent to a finite Ulam-stability problem whichhas a positive solution if the lower-semicontinuous submeasure ϕ : P(N) → [0,∞)giving J as (see [94])

J = X ⊆ N : limn→∞ ϕ(X \ n) = 0is non-pathological. Here are two sample results from this extensive piece of work.

Theorem 14.6 ([31]). Assume MA+TA. Let I and J be two analytic P-idealsgiven by non-pathological submeasures. Then every isomorphism between P (N)/Iand P(N)/J has a completely additive lifting.

Corollary 14.7 ([31]). Assuming MA+TA,

P(N)/I1/n P(N)/I1/√n.

34

The following result shows that the axiomatic assumption of MA+TA is in somesense optimal and points out the close relationship between the axiomatic andmathematical work in this area.

Theorem 14.8 ([31]). The following conditions are equivalent for every pair ofanalytic P-ideals I and J :35

(1) The theory ZFC is insufficient for proving P(N)/I ∼= P(N)/J ;(2) P(N)/I and P(N)/J are not isomorphic via a continuous lifting.(3) The theory ZFC+MA+TA is sufficient for proving P(N)/I P(N)/J ;

Remark 14.9. Following the standard Christensen approach ([18]) to formulatethe notion of non-pathological submeasure in terms of the Fubini property this workwas further refined in [63] allowing the theory of lifting to be extended to a quitelarge class of analytic ideals on N. It should be noted however that there are noknown restrictions on analytic (P-)ideals in the following special case of Problem 1of [122].

Problem 14.10. Suppose Φ : P(N)/I → P(N)/J is an isomorphism between twoquotient algebras over analytic (P-)ideals I and J on N represented by a continuouslifting Φ∗ : P(N)→ P(N). Does Φ has a completely additive lifting?

34Recall that, I1/n = X ⊆ N :∑n∈X

1n<∞, and I1/√n = X ⊆ N :

∑n∈X

1√n<∞.

35The provability here could be taken relative to, say, β-logic, or Ω-logic (see [137]).

28 STEVO TODORCEVIC

Of course, similar questions can be asked for arbitrary isomorphisms assumingTA or one of its natural strengthenings such as PFA or MM. Note that (as alsoindicated in Problem 1 of [122]) in this case one should consider more generalsurjective homomorphisms Φ : P(N)/Fin → P(N)/J . For more about the work inthis direction the reader is referred to the articles [34] and [33].

15. The Open Graph Axiom and the Automorphism RepresentationProblem for the Calkin Algebra

In this section we present recent advances towards the automorphism represen-tation problem in the context of noncommutative algebra or more precisely in thecontext of C*-algebra. So let us recall some standard notation and definitionsfrom this area referring the reader to some standard text such as [80] for furtherbackground.

Definition 15.1. For a separable infinite-dimensional complex Hilbert space H,we associate the following standard objects,

B(H): The C*-algebra of bounded operators on H.K(H) ⊆ B(H): The ideal of compact operatorsC(H) = B(H)/K(H): The Calkin algebra

Recall also that every separable C*-algebra is isomorphic to a subalgebra of B(H).An automorphism Φ of a C*-algebra is inner if there is a unitary u such that

(∀a) Φ(a) = uau∗

It is easily seen that every automorphism of B(H) is inner but the problem in thiscontext is about the following diagram

B(H) Φ∗−−−−→ B(H)

π

y π

yC(H) Φ−−−−→ C(H).

More precisely, one would like to find conditions on representation Φ∗ for a givenautomorphism Φ of the Calkin algebra that guarantee Φ to be inner. In particularone would like to know the answer to the following question.

Question 15.2 ([15]). Is there an outer automorphism of the Calkin algebra?

Theorem 15.3 ([81]). CH implies that there is an outer automorphism of theCalkin algebra.

Theorem 15.4 ([35]). TA implies all automorphisms of the Calkin algebra areinner.

The proof uses several ideas developed in the context of commutative theory (seethe previous section) but it is of course considerably more subtle. We mention onlysome of the new features that appear in the proof. For an orthogonal decomposition~J = (Jn)n<ω of H into finite-dimensional subspaces, let

D( ~J) = a ∈ B(H) : (∀n) a[Jn] ⊆ Jn

K( ~J) = D( ~J) ∩ K(H)

C( ~J) = D( ~J)/K( ~J).


It turns out that an automorphism Φ of the Calkin Algebra C(H) is uniquelydetermined by its restrictions to subalgebras of the form C( ~J) so it is natural tostudy these restrictions. The following result shows that axioms like TA will benecessary in this study.

Theorem 15.5 ([35]). Using CH one can build an automorphism of C(H) that isnot inner but it is inner on C( ~J) for every finite dimensional orthogonal decompo-sition ~J = (Jn) of H.

Fix an automorphism Φ of the Calkin algebra C(H). The proof of Theorem15.4 splits naturally into two parts which resemble the corresponding parts in thecommutative case, the theory of gaps in the quotient structures. In particular thesecond lemma below follows from a more informative fact which shows that underTA every ‘coherent family of unitaries is trivial’ which has its direct analog in thenoncommutative case.

Lemma 15.6 ([35]). TA implies that Φ is inner on C( ~J) for each finite dimensionalorthogonal decomposition ~J = (Jn) of H.

Lemma 15.7 ([35]). TA implies that if an automorphism of C(H) is inner on C( ~J)for each finite dimensional orthogonal decomposition ~J = (Jn) of H then it mustbe inner on C(H).

As in the commutative theory Theorem 15.4 has its ZFC version stated as follows.

Theorem 15.8 ([35]). An automorphism Φ of the Calkin algebra C(H) is inner ifand only if it has a representation Φ∗ : B(H)→ B(H) whose restriction to the unitball of B(H) is measurable relative to the σ-algebra generated by analytic sets.36

16. The Open Graph Axiom and Cofinalities of Banach Spaces

Recall the following well known problem from Banach space theory (see, forexample, [78])

Problem 16.1 (Separable Quotient Problem). Does every infinite-dimensional Ba-nach space X have a separable infinite-dimensional quotient X/Y ?

This problem is related to the following purely set-theoretic notion.

Definition 16.2. The cofinality, cf(X), of a given Banach space X is the minimalordinal γ for which there is an increasing chain Xξ (ξ < γ) of proper closedsubspaces whose union

⋃ξ<γ Xξ is dense in X.

Problem 16.3 (Equivalent Reformulation of the Separable Quotient Problem).Does every infinite-dimensional Banach space have cofinality ω?

Theorem 16.4 ([127]). Assume TA. Then ω and ω1 are the only two possiblevalues for the cofinality cf(X) of a given infinite-dimensional Banach space X.

Sketch of Proof. We consider several cases.Case 1: l1 → X∗. Then applying a combination of a result of Johnson-Rosenthal

and a result of Hagler-Johnson one finds closed Y ⊆ X withX/Y infinite-dimensionaland separable, and therefore cf(X) = ω.

36Here, we take B(H) with its strong operator topology so that its unit ball is Polish.

30 STEVO TODORCEVIC

Case 2: l1 6→ X∗. In this case from the well-known work of about Baire-class-onefunctions([86] and [14]) one can deduce the following property of the dual ball BX∗ :

Every sequence (x∗n) of elements of BX∗ has a convergent subsequence relativeto the weak (and therefore weak*) topology of BX∗ .

In fact, every countable D ⊆ BX∗ is sequentially dense in its closure (even insideBX∗∗∗).

Case 2.1: l1 6→ X∗ and there is countable D ⊆ BX∗ such that D has a non-Gδ-point.

D = fn : n < ω0∗ ∈ D not Gδ relative to D.

A =a ⊆ ω : 0∗ /∈ fn : n ∈ a

B = A⊥ = a ⊆ ω : (∀a ∈ A) b ∩ a ∈ Fin

A0 =a ⊆ ω : (fn)n∈a converges to a point 6= 0∗

Note 16.5. A0 is a dense subset of B, i.e., for every infinite b ∈ B there is infinitea ∈ A0 such that a ⊆ b.

Claim 1. No countable subset of P(ω) separates A and B, or equivalently, nocountable subset of P(ω) separates A0 and B.

So applying TA, we get a biorthogonal gap (aξ, bξ) : ξ < ω ⊆ A0×B. For ξ < ω1,let

gξ = limn∈aξ

fn.

Finally, for α < ω1, let

Xα = x ∈ X : (∀ξ ≥ α) gξ(x) = 0 .Then Xα $ X, Xα ⊆ Xβ for α < β < ω1, and X =

⋃α<ω1

Xα. It follows thatcf(X) ≤ ω1.

Case 2.1*: l1 6→ X∗ and there is countable D ⊆ BX∗ whose closure contains aclosed non-Gδ-subset F . Set

A =a ⊆ ω : (fn)n∈a converges to a point of F

B =

b ⊆ ω : (fn)n∈b converges to a point outside F

Claim 2. No countable subset of P(ω) separates A and B.

Applying again TA, we get a biorthogonal gap (aξ, bξ) : ξ < ω1 ⊆ A× B. Forξ < ω1, let

gξ = limn∈aξ

fn, hξ = limn∈bξ

fn

andeξ = gξ − fξ(6= 0∗!).

For α < ω1, letXα = x ∈ X : (∀ξ ≥ α) eξ(x) = 0 .

Then Xα 6= X, Xα ⊆ Xβ for α < β < ω1, and X =⋃α<ω1

Xα. So again, we getthat cf(X) ≤ ω1.

Case 2.2: There is countable D ⊆ BX∗ whose closure K = D is not second-countable though all of its closed subsets are relatively Gδ. For f ∈ K, let Vn(f)


(n < ω) be a fixed neighborhood base of f in K. Then using the assumption thatK is not metrizable we can find a sequence (fα, gα) α < ω1 of pairs of distinctelements of K such that if α < β and n < ω then the open sets Vn(fα) and Vn(gα)do not separate the points fβ and gβ . Let

hβ = fβ − gβ (β < ω1).

Then one can easily check that

(∀x ∈ X) β < ω1 : hβ(x) = 0 is countable.

So, as before, if for α < ω1, we set

Xα = x ∈ X : (∀β ≥ α) hβ(x) = 0

we get that Xα 6= X, Xα ⊆ Xβ for α < β < ω1, and X =⋃α<ω1

Xα. It followsthat cf(X) ≤ ω1 also in this case.

Case 2.3: For every countable D ⊆ BX∗ , the closure KD = D is second-countable. Choose an increasing sequence Dα (α < ω1) of countable subsets ofBX∗ such that

Dα $ Dα+1 for all α < ω1.

LetK =

⋃α<ω1

Dα.

Case 2.3.1: K 6=⋃α<ω1

Dα. Then no point of K \⋃α<ω1

Dα is in the closure ofa countable subset of

⋃α<ω1

Dα, so by ([58]), we could find a sequence fξ : ξ <ω1 ⊆

⋃α<ω1

Dα converging to a point g ∈ K \⋃α<ω1

Dα. Let

hξ = fξ − g (ξ < ω1).

Then as before, if for α < ω1, we set

Xα = x ∈ X : (∀ξ ≥ α) hξ(x) = 0


Xα. It followsthat cf(X) ≤ ω1 in this case as well.

Case 2.3.2: K =⋃α<ω1

Dα. By our assumption each Dα is second countable,so the compactum K, while non-metrizable, has a basis of size ℵ1. So fix a basisVξ (ξ < ω1) for K consisting of co-zero subsets of K. Then we can find a sequence(fα, gα) (α < ω1) of pairs of disctinct elements of K such that

(∀ξ < α < ω1) fα ∈ Vξ ⇔ gα ∈ Vξ.

Lethα = fα − gα(α < ω1).

Since Vξ (ξ < ω1) is a basis for K, it follows that

(∀x ∈ X) α < ω1 : hα(x) = 0 is countable.

So if for α < ω1, we set

Xα = x ∈ X : (∀β ≥ α) hβ(x) = 0


Xα. So we getcf(X) ≤ ω1 in this case as well.

32 STEVO TODORCEVIC

17. The Open Graph Axiom and the Theory of Analytic Gaps

The axiom TA is one of those set theoretic principles that can sometimes beremoved from intended applications giving us results that do not rely on any ad-dition axiom at all. In this section we give some examples of results obtained thisway. They concern the theory of definable gaps in the quotient structures such asP(N)/Fin or NN/Fin.

Example 17.1. Let Seq be the tree of finite sequences of integers ordered byend-extension. Let CHSeq be the family of all chains of Seq and let ISSeq be thefamily of subsets of Seq that are immediate successors of some node of Seq, i.e., thedownwards closure of the family tan : n ∈ N, (t ∈ Seq).

Note that CHSeq and ISSeq are two orthogonal analytic families of subsets of thetree Seq which can’t be separated and, in fact, the family CHSeq of chains is notcountably generated in the orthogonal (ISSeq)⊥. The following result shows thatthis is a canonical object in this category of gaps.

Theorem 17.2 ([118]). Suppose that A and B are two orthogonal downwards closedfamilies of subsets of N and that A is analytic. Then either,

(1) A is countably generated in B⊥, or(2) there is a 1-1 map ϕ : Seq→ N such that ϕ”CHSeq ⊆ A and ϕ”ISSeq ⊆ B.

Corollary 17.3 ([118]). If an analytic ideal I on N has the diagonal sequenceproperty37 then its orthogonal I⊥ is countably generated.

Proof. We show that I⊥ is an Fσ ideal and then the conclusion follows by applyingthe theorem to the pair (I⊥, I) of analytic orthogonal families.38 Let

X = K ∈ K(2N) : (∃b ∈ I)(∀x ∈ K) b ∩ x /∈ FIN.

Note that A is an analytic subset of the space K(G) of compact subsets of G =P(N)\I⊥. Since we need to prove that G is a Gδ subset of the Cantor cube, applyingChristensen’s theorem ([17]), it suffices to show that every countable compact subsetK of G belongs to X . For each x ∈ K fix ax ∈ I such that x ∩ ax is infinite. SinceK is countable, and since I has the diagonal sequence property, there is b ∈ I suchthat x∩ ax ∩ b is infinite for all x ∈ K. It follows that b∩ x is infinite for all x ∈ K.This shows that K ∈ X .

Corollary 17.4 ([118]). Suppose I and J are two orthogonal P-ideals on N. Ifone of them is analytic then they can be separated.

Remark 17.5. So Hausdorff (ω1, ω1)-gaps in P(N)/Fin (see, [48],[49]) are highlynon-analytic!

Let us now examine the case when both of the two orthogonal families are defin-able, or in other words, look for the symmetric version of the analytic gap theorem.

37An ideal I has the diagonal sequence property if for every an : n < ω ⊆ I \ Fin there is

b ∈ I such that b ∩ an 6= ∅ for all n < ω. Clearly, every P-ideal has this property.38In fact, we only need the consequence saying that if A and B are orthogonal ideals such that

A is analytic and B has the diagonal sequence property then A is countably generated in B⊥.


Example 17.6. Let Seq2 be the tree of finite sequences of 0’s and 1’s. We identifythe Cantor space 2N with the collection of all infinite branches of this tree. Forx ∈ 2N and i < 2, let

aix = t ∈ Seq2 : t v x and x(|t|) = i.Fir i < 2, let AiSeq2

= aix : x ∈ 2N. Then A0Seq2

and A1Seq2

are two orthogonalanalytic (in fact, perfect) families of chains of the tree Seq2 which form a gap whichis not even countably separated. In fact using our terminology from Section 13above, A0

Seq2and A1

Seq2form a perfect biorthogonal gap.

The following result shows that this is a canonical object in this category ofanalytic gaps.

Theorem 17.7 ([118]). Suppose that A and B are two orthogonal analytic familiesof subsets of N closed under finite changes. Then either,

(1) A and B are countably separated, or(2) there is a 1-1 map ϕ : Seq2 → N such that ϕ”A0

Seq2⊆ A and ϕ”A1

Seq2⊆ B.

A generalization of this result to more than two orthogonal analytic families isgiven in [4]. Let us mention a typical application of this result.

Theorem 17.8 ([123]). Suppose K is a separable compact subset of Baire-class-1 functions on some complete metric space X which in the topology of pointwiseconvergence on X has the constant zero function 0 as a non-Gδ point. Then thereis a 1-1 Souslin-measurable map ϕ : 2N → K whose range is weakly null.39

We mention an interesting recent application of this result to the separable quo-tient problem.

Theorem 17.9 ([3]). Every dual Banach space X has a separable quotient.

Sketch. Let X = Y ∗ and replacing X and Y we my assume Y is separable and thatX is not. We may also assume that `1 does not embed into Y since otherwise `∞would be a quotient of X. Thus the unit ball BY ∗∗ = BX∗ is a separable compact setof Baire-class-1 functions on X (see [79]) with 0∗∗ a non-Gδ point. So Theorem 17.8applies giving us a 1-1 Souslin-measurable ϕ : 2N → BX∗ whose range is weakly null.We may assume that the range of φ is semi-normalized. Now note that for everypositive integer k and C > 1 the set of all 1-1 sequences (ai)i<k of elements of 2N

for which the corresponding sequence (φ(ai))i<k is C-unconditional40 is comeagerin (2N)k. So by Mycielsky’s theorem ([129]; 6.40), we get a perfect unconditionalsequence and in particular an infinite unconditional sequence in X∗. So by theresults of Johnson and Rosenthal ([57], [45]), X has a separable quotient.

As the reader could have already guessed, if D is a countable dense subset ofthe compactum K from the hypothesis of Theorem 17.8 one obtains φ from thealternative (2) of Theorem 17.7, where A is the set of all subsets of D which don’taccumulate to 0 and where B is its orthogonal, i.e., the set of all subsets of D whichconverge to 0. It is not hard to see that (1) fails for the gap formed by A and B.Note however that Theorem 17.7 does not quite apply here since B is not analytic.

39In other words, for every x ∈ X and ε > 0 the set a ∈ 2N : |φ(a)(x)| ≥ ε is finite.40Recal that a sequence (xi)i∈I of elements of some normed space (N, ‖ · ‖) is C-unconditional

if ‖∑i∈F xi ‖≤ C ‖

∑i∈G xi ‖ for every pair F ⊆ G of finite subsets of I.

34 STEVO TODORCEVIC

Examining the proof of Theorem 17.8 one discovers the reason of why we still havethe alternative (2) for these A and B and gets to the following result.

Theorem 17.10 ([24]). Suppose that A and B are two orthogonal families of sub-sets of N closed under finite changes, that A is analytic, that B is Souslin-measurableand that B has the weak diagonal sequence property.41 Then either,

(1) A and B are countably separated, or(2) there is a 1-1 map ϕ : Seq2 → N such that ϕ”A0

Seq2⊆ A and ϕ”A1

Seq2⊆ B.

Applications of the gap dichotomies to the study of diagonal sequence propertiesin the class of analytic topological spaces is given in [130].

18. Open Graphs on Polish Spaces

In this section we examine this phenomenon in some details by considering thefollowing abstract property drown from the statement of TA.

Definition 18.1. Given a topological space X we let TA(X) denote the statementthat for every open graph G = (X,R) on X either

(1) Chr(G) ≤ ℵ0, or(2) Kℵ1 ≤ G.

The point of this definition is to isolate spaces X for which TA(X) is simply true.For example, we have the following simple result.42

Proposition 18.2. TA(X) is true for X = NN.

Moreover, we have the following preservation result for this property.

Lemma 18.3. TA(X) implies TA(Y ) whenever Y is a continuous image of X.

Proof. Fix a continuous onto map f : X → Y and an open graph G = (Y,R) on Y .Define

S =

(x, y) ∈ X2 : f(x) 6= f(y) & (f(x), f(y)) ∈ R.

Then H = (X,S) is an open graph on X and an application of TA(X) to H givesus the conclusions of TA(Y ) for the graph G.

Corollary 18.4. TA(X) is true for every analytic space X, and so in particular,for every Polish space X.

Remark 18.5. In fact, the second alternative of TA(X) can be strengthened asfollows:

(2)∗ K2N ≤c G.Let us denote this strong form of TA(X) by TA*(X). It turns out that TA*(X)for X belonging to some class Γ of separable metric spaces can naturally be consid-ered as the strengthening of the standard perfect-set property of Γ (that is, everyuncountable X ∈ Γ contains a perfect set). The following results shows that thereis indeed a quite close relationship between this new perfect-set property and theclassical one.

41We say that an ideal I has the weak diagonal sequence property of for every sequence an :n < ω ⊆ I \ Fin there is b ∈ I such that b ∩ an 6= ∅ for infinitely many n’s.

42The reader is indeed invited to reproduce the simple direct proof of Proposition 18.2 or relyon Shoenfield’s absoluteness and the fact that the second alternative of TA(X) can be forced ifthe first fails (see Proposition 18.8 below). More proofs of this result can be found in [37].


Theorem 18.6 ([37]). The following statements are equivalent:(1) Every uncountable co-analytic set of reals contains a perfect set.(2) TA*(X) holds for every co-analytic set of reals.(3) For every real r, L[r] fails to correctly compute the first uncountable ordinal.

Remark 18.7. Recall that the equivalence of (1) and (3) is a classical result ofSolovay [96].

The following fact (due to the author and appearing in [37]) gives us an expla-nation of the close relationship between the one-dimensional and two-dimensionalperfect-set properties, in view of the fact that we can always go to a forcing exten-sion of the universe satisfying TA and p > ω1.

Proposition 18.8. Assume p > ℵ1. Suppose G = (X,R) is an open graph on aPolish space X and that Y is an arbitrary subset of X. The following are equivalent:

(1) There is an uncountable G-complete subset Z ⊆ Y ,(2) There is a G-complete Gδ-set G such that G ∩ Y is uncountable.

Proof. To prove the nontrivial implication, we may assume that Z is ℵ1-dense inX. Fix a countable basis B of X. Consider the set P of all mappings p : Zp → Bwhere Zp is a finite subset of Z such that x ∈ p(x) for all x ∈ Zp and

(∀x 6= y ∈ Zp) p(x) ∩ p(y) = ∅ and p(x)× p(y) ⊆ R.We order P by letting p ≤ q whenever Zp ⊇ Zq and the family p′′Zp refines thefamily q′′Zq. Then P is a σ-centered poset, so our assumption p > ℵ1 gives usa filter F of P such that

⋃p∈F Zp is uncountable and such that if we let G =⋂

p∈F⋃p′′Zp then G is the desired G-complete G-delta set whose intersection with

Z is uncountable.

Remark 18.9. This also shows (using the same assumption) that whenever a suffi-ciently rich point-class Γ has the Perfect-Set Property then TA(Γ) implies TA∗(Γ).

We finish this section with a typical application of TA∗.

Theorem 18.10 ([120]). If X is a Polish space and if K is any compact spacethen the family of all homeomorphic copies of K inside X is σ-linked43 unless theproduct K × 2N embeds into X.

19. Chromatic Theory of Closed Graphs on Polish Spaces

In this section we show the difference between the classes of open and closedgraphs on separable metric spaces.

Example 19.1 ([113]). [The closed graph Gc = (NN, Rc)] To define Rc we firstassociate to every f ∈ NN a sequence fi as follows. Let n0 < n1 < · · · be the listof all n such that f(2n+ 1) 6= 0. For a given i let fi be defined by

fi nk = f nk and fi(nk + j) = f(2i(2j + 1))

for j < ω, where k = min` : f(2n0 +1)+ · · ·+f(2n`+1) > i; let fi = f wheneversuch k does not exist. Finally, define

(f, g) ∈ Rc ⇐⇒ (∃i) [f = gi or g = fi] .

43Recall that a family F of nonempty sets is σ-linked if there is a countable decompositionF =

⋃n<ω Fn such that (∀n < ω)(∀F,G ∈ Fn) F ∩G 6= ∅.

36 STEVO TODORCEVIC

Claim 1. Rc ∪4 is a closed symmetric relation on NN.

Claim 2. Kℵ1 Gc =(NN, Rc

).

Claim 3. Chr (Gc) = ℵ1.

The following fact will give us Theorem 11.4 showing that in the context offorcing axiom the compactness number κC has only the trivial realization κC = c+.

Proposition 19.2 ([108]). Under MAℵ1 , every subgraph of the graph Gc of size atmost ℵ1 is countably chromatic.

On the other hand the following result shows that there are natural conditionsunder which G does have subgraphs of size ℵ1 that are not countably chromatic.

Proposition 19.3 ([115]). If b = ℵ1 there is X ⊆ NN of size ℵ1 such that the

restriction Gc X is not countably chromatic and in fact contains no uncountableGc-complete set.

Question 19.4. Is b > ℵ1 equivalent to the statement that all subgraphs of Gc ofsize ℵ1 are countably chromatic?

The example Gc = (NN, Rc) is a closed graph on a Polish space which leads usto the following natural question.

Question 19.5. Is the set of cardinals

1, 2, 3, . . . ∪ ℵ0,ℵ1, cequal to the chromatic number spectrum of closed graphs on Polish spaces?

It is known that the positive answer to this question is consistent so the question isreally about a possibility of resolving this question using no additional axioms. Toexplain this recall that a graph G = (X,E) satisfies the countable chain condition ifthe poset of finite G-independent subsets of X satisfies the countable chain condi-tion, or in other words if any uncountable family F of finite G-independent subsetsof X contains two distinct sets p and q whose union p ∪ q is also G-independent.The following dichotomy result shows that among closed graphs on Polish spacesthere are many examples of graphs satisfying the countable chain condition.

Theorem 19.6 ([113]). For every closed graph G = (X,E) on a Polish space Xeither,

(1) G satisfies the countable chain condition, or(2) K2N ≤c G.

Proof. Assume K2N 6≤c G and, towards showing that G satisfies the countable chaincondition, let F be an uncountable family of finite G-independent subsets of X.Applying a ∆-system argument one sees that we may assume that the elements ofF are pairwise disjoint and of some fixed cardinality n. Moreover, we may assumethat for a fixed sequence B0, ..., Bn−1 of open sets with pairwise disjoint closureswe have that every element of F meets every Bi in exactly one point and that

(∀i 6= j < n) (Bi ×Bj) ∩ E = ∅.Thus F can be naturally identified with a subset of the product

∏i<nBi. Let Y be

the closure of F in this product. If there exist x, y ∈ Y such that

(∀i < n)[xi 6= yi and (xi, yi) 6∈ E],


using the fact that F is dense in Y, we would find p 6= q in F such that p ∪ q isG-discrete, as required. So, we assume that such x and y cannot be found in Y, andwork for a contradiction. Choose a perfect subset P of Y such that

(∀x 6= y ∈ P )(∀i < n) xi 6= yi.

Thus, for every x 6= y in P there is i < n such that (xi, yi) ∈ E. Applying Galvin’sperfect-set Ramsey theorem (see [13]), we find a perfect Q ⊆ P and i < n such that

(∀x 6= y ∈ P ) (xi, yi) ∈ E.

It follows that the projection of Q on the ith coordinate is a perfect G-completesubset of X, a contradiction.

Recall also the following statement considered above in Section 9.

Definition 19.7 ([112]). [The Principle Σ2B ] Chr(G) ≤ ℵ1 for every Borel graph G

satisfying the countable chain condition.

It has been shown in [112] that Σ2B is consistent with the continuum large though

it is still open whether the unrestricted version Σ2 equivalent to CH? Returningback to our problem of chromatic spectra of closed graphs on Polish spaces, wehave the following result (an immediate consequence of Theorem 19.6) that showsthat the minimal value of the chromatic spectrum is at least consistent with theusual axioms of set theory.

Theorem 19.8 ([113]). Assuming Σ2B , for every closed graph G = (X,E) on a

Polish space X, either(1) Chr(G) ≤ ℵ1, or(2) K2N ≤c G.

Corollary 19.9. Σ2B implies that the chromatic number Chr(G) of a closed graph

G = (X,E) on a Polish space X must take one of the following possible values1, 2, 3, . . . , ℵ0,ℵ1, c.

What we were really showing so far is the extent to which the complete graphG2N is a critical object in the class of all closed graphs on Polish spaces. We nowgive another similar example of a closed graph which itself is a critical object inthis class though relative to the Borel chromatic theory rather that the ordinarychromatic theory of graphs.

Example 19.10 ([64]). [The Closed Acyclic Graph G0 = (2N, R0)] Choose a se-quence tn ∈ 2n (n < ω) such that (∀t ∈ 2<ω) (∃n) t v tn. For x ∈ 2ω we nowdefine (xi)i<ω ⊆ 2ω as follows. Let n0 < n1 < n2 < · · · be the list of all n suchthat x n = tn. Let

xi ni = x ni

xi (ni) = 1− x (ni)

xi(n) = x(n) for n > ni

whenever ni exists; otherwise let xi = x. Finally put

(x, y) ∈ R0 ⇐⇒ (∃i) [x = yi or y = xi] .

Claim 1. G0 = (2N, R0) is a closed acyclic graph and therefore Chr(G0) = 2.

38 STEVO TODORCEVIC

However, the corresponding Borel chromatic number and the correspondingBaire chromatic number of G0 behave quite differently.

Claim 2. ChrB (G0) > ℵ0, or in other words, for every countable Borel partition,and in fact, for every countable Baire partition,

2N =⋃i<ω Bi

there exists i < ω and x 6= y in Bi such that (x, y) ∈ R0.

The following result shows that G0 is the minimal Borel graph of uncountableBorel chromatic number in a similar way the complete graph Kℵ1 is minimal in theclass of open uncountably chromatic graphs on separable metric spaces.

Theorem 19.11 ([64]). Let G = (X,R) be an analytic graph 44 on a Polish spaceX. Then exactly one of the following holds:

(1) ChrB(G) ≤ ℵ0,(2) G0 ≤c G. 45

Corollary 19.12 ([93]). A co-analytic equivalence relation E defined on a Polishspace X has either countably many classes or a perfect set of pairwise inequivalentelements.

Proof. Apply Theorem 19.11 to the analytic graph G = (X,X2 \ E).

We finish this section with a remark that many new applications and variationsof the G0-dichotomy can be found in the recent article [74].

Part 3. Ideals of Countable Sets

20. Two Dual Dichotomies for Ideals of Countable Sets

In this section we introduce two dichotomies for ideals of countable subsets ofsome index set S. There are two conditions that one needs to put on such ideals inorder to obtain consistent dichotomies.

Definition 20.1. Suppose I ⊆ [S]≤ℵ0 is an ideal 46. I is a P-ideal if for everysequence an : n < ω ⊆ I there exists b ∈ I such that an ⊆∗ b 47 for all n < ω.

Example 20.2. The ideal I = [S]≤ℵ0 of all countable subsets of some set S is aP-ideal.

Example 20.3. For C ⊆ P(S) countable (or, more generally, for C ⊆ P(S) of size< b), the orthogonal

I = C⊥ = a ∈ [S]≤ℵ0 : (∀C ∈ C) a ∩ C ∈ Finis a P-ideal.

So the larger cardinal b is, the larger is the class of P-ideals, and therefore strongerare the dichotomies that we would like to associate to this class of ideals. Thefollowing simple fact is showing us how the operation of taking the orthogonaltransfers the dichotomies that one usually would like to consider in this context.

44that is, R ⊆ X2 is analytic45i.e., there is a continuous homomorphism Φ :

(2N, R0

)→ (X,R) .

46A collection of subsets of S closed under taking unions and subsets and x : x ∈ S ⊆ I.47Recall that an ⊆∗ b means that an \ b ∈ Fin.


Proposition 20.4. Let J be a < p-generated ideal of countable subsets of some setS and let I = J⊥ be its orthogonal. Then the following statements are equivalent:

(A) The ideal J has one of the following two properties:(1) There is uncountable X ⊆ S such that X ∩ b ∈ Fin for all b ∈ J .(2) There is a decomposition S =

⋃n<ω Sn such that [Sn]ℵ0 ⊆ J for all

n < ω.(B) The ideal I has one of the following two properties:

(1) There is uncountable X ⊆ S such that [X]ℵ0 ⊆ I.(2) There is a decomposition S =

⋃n<ω Sn such that Sn ∩ a ∈ Fin for all

n < ω and a ∈ I.

The dichotomy (A) appears in [106] in notation (*) as the combinatorial essenceof several problems solved by the method of taking elementary submodels as sideconditions when building proper posets (see [102],[104],[108]).

Theorem 20.5 ([106]). Assume PFA. Then for every ℵ1-generated ideal J ofcountable subsets of some set S, either

(1) there is uncountable X ⊆ S such that X ∩ b ∈ Fin for all b ∈ J , or(2) there is a decomposition S =

⋃n<ω Sn such that [Sn]ℵ0 ⊆ J for all n < ω.

The proof consists in showing that if (2) fails then there is a proper poset thatforces the uncountable set satisfying (1). Looking back at the Proposition 20.4,we see that a proper poset would force alternative B(1) whenever B(2) fails forevery P-ideal I (not just one of the form I = J⊥)48. That the proof of Theorem20.5 gives the second dichotomy (B) with no restriction on the ideal I (except thatit is a P-ideal) was realized by the author while solving specific problems aboutcoherent sequences on ω1 (see, for example, [102],Theorem 6, [108],Theorem 8.13,[115], Lemma 1 and Lemma 1*, and [30]) asked by Galvin [43], Dow [25], Leidermanand Malykhin [68], and Watson (Problem 174 from [75] 49).

Theorem 20.6 ([102],[108],[115],[30]). Assume PFA. Then for every P-ideal I ofcountable subsets of some set S, either

(1) there is uncountable X ⊆ S such that [X]ℵ0 ⊆ I, or(2) there is a decomposition S =

⋃n<ω Sn such that Sn ∩ a ∈ Fin for all n < ω

and a ∈ I.

[Sketch of Proof] Fix a P-ideal I of countable subsets of S and assume that thesecond alternative fails, or in other words, S cannot be decomposed into count-ably many sets orthogonal to I. We shall find a proper poset P which forces anuncountable set X ⊆ S such that [X]ℵ0 ⊆ I.

Fix a sufficiently large regular cardinal θ such that the structure (Hθ,∈) containsthe set [S]ℵ0 . For a countable elementary submodel N ≺ Hθ, we fix bN ∈ I suchthat bN ⊆ S ∩N and

(∀a ∈ I ∩N) a ⊆∗ bN ,and we fix a point xN ∈ S such that

(∀X ∈ P(S) ∩N)[X ⊥ I ⇒ xN /∈ X].

48Note that by Example 20.3, the ideal I of Proposition 20.4 is indeed a P-ideal.49The problem was motivated by the well known problem of the existence of Ostaszewski space

on the basis of CH rather than ♦.

40 STEVO TODORCEVIC

Expanding (Hθ,∈) by adding a well-ordering <w of Hθ we may assume N 7→ bNand N 7→ xN are definable maps.

Let P be the collection of all finite ∈-chains p of countable elementary submodelsof (Hθ,∈, <w). For p, q ∈ P, set p ≤ q if and only if,

(a) p ⊇ q, and(b) (∀M ∈ q)(∀N ∈M ∩ (p \ q))[xN ∈ bM ].

Claim 1. P is a proper poset.

Proof Sketch. Consider a sufficiently large regular cardinal κ and a countable ele-mentary submodel M ≺ (Hκ,∈) containing P, and let p ∈ P ∩M be arbitrarilychosen. Let

q = p ∪ M ∩Hθ .We shall show that q is (M,P)-generic. Consider dense-open D ⊆ P, D ∈M , andr ≤ q. We may assume r ∈ D. For s ∈ D end-extending r ∩M , let

xs = 〈xs1, xs2, . . . , xsl 〉

wherexN : N ∈ s \ (r ∩M) = xs1, xs2, . . . , xsl

enumerated according to the ∈-ordering of s. Let k be the length of the sequencexr associated to the condition r which clearly end-extends r ∩M. Let

F =x ∈ Sk : (∃s w r ∩M) [s ∈ D & xs = x]

.

Then F ∈M and xr = 〈xN1 , xN2 , . . . , xNk〉 ∈ F . It follows that F is Namba relativeto the co-ideal

H = X ⊆ S : I X 6|= alternative (2).So in particular F contains a subfamily F0 ∈ M that forms the set of maximalnodes of an everywhere H-branching subtree of S≤k of height k. Let

X1 = ξ ∈ S : (∃x ∈ F0) ξ = x1 .

Then X1 ∈ M ∩ H. So there is infinite a1 ∈ I ∩M such that a1 ⊆ X1. Thena1 ⊆∗ bNi for all i = 1, 2, . . . , k. So we can pick ξ1 ∈ a1 such that ξ1 ∈

⋂ki=1 bNi .

LetX2 = ξ ∈ S : (∃x ∈ F0) [ξ1 = x1 & ξ = x2] .

Then X2 ∈M ∩H. So there must be infinite a2 ∈ I ∩M such that a2 ⊆ X2. Thenagain a2 ⊆∗

⋂ki=1 bNi , so we can choose ξ2 ∈ a2 ∩

⋂ki=1 bNi , and so on. At the end

of this process one obtains ξ1, ξ2, . . . , ξk ⊆⋂ki=1 bNi such that

〈ξ1, ξ2, . . . , ξk〉 ∈ F ∩M.

Pick r ∈ D ∩M end-extending r ∩M such that xr = 〈ξ1, ξ2, . . . , ξk〉 . Then r and rare two compatible conditions of P and, in fact, r∪r is their common extension.

It turns out that the second dichotomy (B) is a statement with an interestindependent from PFA, so let us give it a special name.

Definition 20.7. [P-Ideal Dichotomy] For every P-ideal I of countable subsets ofsome set S, either

(1) there is uncountable X ⊆ S such that [X]ℵ0 ⊆ I, or(2) S can be decomposed into countably many sets orthogonal to I.


The following result is only one of the facts which points out the special characterof the second dichotomy.

Theorem 20.8 ([124]50). The P-ideal dichotomy is consistent with the GCH rela-tive to the consistency of the existence of a supercompact cardinal.

21. PID and the Hausdorff Condition on Gaps

Fix an infinite set S and for each countable subset a of S fix a well ordering <aof a of order type ≤ ω. Then for n < ω, we can let a[n] denote the set formed bytaking the first n elements of a according to <a . If S itself is countable, it is moreconvenient to fix its enumeration S = sk : k < ω and for a ⊆ S and n < ω, leta[n] = a ∩ sk : k < n. This will give us a way to state the Hausdorff originalcondition on gaps in the context of P(S)/Fin rather than just P(N)/Fin.

Definition 21.1. A sequence (aα, bα) : α < ω1 of pairs of countable subsets ofS satisfies the Hausdorff condition if

bα ⊆∗ bβ whenever α < β < ω1,(1)

aα ∩ bβ ∈ Fin for all α, β < ω1, and(2)

α < β : aα ∩ bβ ⊆ aα[n] ∈ Fin for all β < ω1 and n < ω(3)

Remark 21.2. In Hausdorff’s original definition, the set S was countable and thesequence aα (α < ω1) was also assumed to be increasing under ⊆∗ . Note also thatthis notion does not depend on the choice of well-orderings <a (a ∈ [S]≤ℵ0).

First in order is a lemma which shows that a sequence satisfying the Hausdorffcondition is indeed a gap in a quite strong and absolute sense.

Lemma 21.3. Suppose that a sequence (aξ, bξ) : ξ < ω1 of pairs of countablyinfinite subsets of some set S satisfies the Hausdorff condition. Then aξ : ξ < ω1is not countably generated in the orthogonal of bξ : ξ < ω1.

Proof. Otherwise we can find a single subset C of S and an uncountable subset Γof ω1 such that aγ ⊆∗ C and bγ ∩C ∈ Fin for all γ ∈ Γ. We may assume that thereis some integer n such that

aα \ C ⊆ aα[n] for all α ∈ Γ.

Moreover, increasing n and shrinking Γ, we may assume that the sequence of finitesets b∗β = bβ ∩ C (β ∈ Γ) forms a ∆-system with root r such that

aα ∩ r ⊆ aα[n] for all α ∈ Γ, and

aα ∩ (b∗β \ r) = ∅ for all α, β ∈ Γ with α < β.

Choose β ∈ Γ such that Γ0 = Γ ∩ β is infinite. Then aα ∩ bβ ⊆ aα[n] for all α ∈ Γ0

contradicting the Hausdorff condition.

Fix two orthogonal families A and B of countable subsets of S. Let

I(A,B) = A ∈ [A]≤ℵ0 : (∃b ∈ B)(∀n < ω) a ∈ A : a ∩ b ⊆ a[n] ∈ Fin.

50The special case S = ω1 of this result appears in [2] and it does not require large cardinals.However, as we are going to see below, even the case S = ω2 has some large cardinal strength.

42 STEVO TODORCEVIC

Lemma 21.4. I(A,B) is a P-ideal of countable subsets of A whenever B is a P-idealof countable subsets of S and either S is countable or A consists of pairwise disjointsubsets of S.

Proof. We only treat the assumption that A consists of pairwise disjoint sets sincethe treatment of the other assumption is quite analogous and simpler. So letAk : k < ω ⊆ I(A,B) be a given sequence and for each k, let bk ∈ B witnessAk ∈ I(A,B). Pick b ∈ B such that bk ⊆∗ b for all k < ω. Note that for eachn, k < ω, the set

Ak(n) = a ∈ Bk : a ∩ b ⊆ a[n]is finite, since

Ak(n) ⊆ a ∈ Ak : a ∩ bk ⊆ a[n] ∪ a ∈ Ak : a ∩ (bk \ b) 6= ∅.Let A =

⋃n<ω(An \An(n)). Then b is a witness of A ∈ I(A,B). Moreover, An ⊆∗ A

for all n < ω, as required.

Lemma 21.5. Suppose that A and and B are two orthogonal ideals of countablesubsets of some set S and that B is a P-ideal. If there is uncountable X ⊆ A suchthat [X ]ℵ0 ⊆ I(A,B) then there is a sequence (aξ, bξ) : ξ < ω1 ⊆ A× B satisfyingthe Hausdorff condition.

Proof. Choose a one-to one sequence aα : α < ω1 ⊆ X . Then for every β < ω1,we can chose bβ ∈ B witnessing that aα : α < β ∈ I(A,B). During the course ofthe proof of Lemma 21.4 we have seen that any set which almost includes bβ willstill witness this, so we may further assume that bβ ⊆∗ bγ whenever β < γ < ω1.Then (aα, bα) : α < ω1 satisfies the Hausdorff condition.

Lemma 21.6. Suppose that A and B are two orthogonal ideals of countable subsetsof some set S and that B is a P-ideal. If A can be covered by countably many setsAn (n < ω) orthogonal to I(A,B), then A is countably generated in B⊥.

Proof. If S is countable, take C = ⋃An : n < ω. Then C ⊆ B⊥ and every element

of A is almost included in some set from C. Assume now that A consists of pairwisedisjoint sets. Fix n < ω. Then for every b ∈ B there is k(b, n) < ω such that

a ∩ b ⊆ a[k(b, n)] for all a ∈ Ak.Since B is σ-directed under ⊆∗ there is k(n) < ω such that b ∈ B : k(b, n) = k(n)is ⊆∗-cofinal in B. Let C =

⋃a\a[k(n)] : a ∈ An : n < ω. Then C is a countable

family of subsets of S that are orthogonal to B such that every element of A isalmost included in some set from C

.

Theorem 21.7 ([124]). Assume PID. Let A and B be two orthogonal families ofsubsets of some countable set S such that the family B is a P-ideal. Then either

(1) A× B contains a Hausdorff gap (aα, bα) : α < ω1, or(2) A is countably generated in B⊥.

Corollary 21.8. Assuming PID, the orthogonal of any Pℵ1-ideal 51 on ω is count-ably generated.

51Recall that an ideal I of countable subsets of some set is a Pℵ1 -ideal if for every sequence

aξ : ξ < ω1 ⊆ I there is b in I such that aξ \ b is finite for all ξ < ω1.


Corollary 21.9. PID implies b ≤ ℵ2.

Question 21.10. Does PID imply c ≤ ℵ2?

Theorem 21.11 ([108]). Assume PID. Let A and B be two orthogonal families ofcountable subsets of some set S such that the family B is a P-ideal whose orthogonalB⊥ is ℵ1-generated. Then either

(1) A× B contains a Hausdorff gap, or(2) A is countably generated in B⊥.

Proof. If there is countable S0 such that A S0 is not countably generated in theorthogonal of B S0, we are done by Theorem 21.7. Otherwise, assuming that A isnot countably generated in B⊥, and using the fact that B⊥ = X ⊆ S : X ⊥ B isan ℵ1-generated ideal in P(S)/Fin, we can find a subfamily A0 of A consisting ofpairwise disjoint sets and still not countably generated in B⊥. Then using Lemmas21.4, 21.5 and 21.6, and PID, we get a Hausdorff gap (aα, bα) : α < ω1 inA0 × B.

Corollary 21.12. Assume PFA. Let A and B be two orthogonal families of count-able subsets of some set S such that the family B is a P-ideal. Then either

(1) A× B contains a Hausdorff gap, or(2) A is countably generated in B⊥.

Proof. This is really the generic version of Theorem 21.11 as it can be reducedto it by noticing that if B is a P-ideal of countable subsets of some set S thenthe orthogonal B⊥ computed in any σ-closed forcing extension is generated by theorthogonal B⊥ computed in the ground model.

22. PID and Coherent Sequences

Definition 22.1. Two maps f : a→ ω and g : b→ ω are coherent if

x ∈ a ∩ b : f(x) 6= g(x) ∈ Fin.

Definition 22.2. Two maps f : a→ ω and g : b→ ω are weakly coherent if

(∀c ⊆ a ∩ b) [f c is unbounded ⇐⇒ g c is unbounded] .

Theorem 22.3 ([124]). Assume PID. Then for every weakly coherent family

fa : a→ ω (a ∈ I)

indexed by some P-ideal I on some set S, either(1) there is uncountable X ⊆ S such that [X]ℵ0 ⊆ I and fa X is finite-to-one

for all a ∈ I, or(2) there is g : S → ω which coheres weakly with all fa (a ∈ I).

Proof. Let

I0 = a ∈ I : |a| ≤ ℵ0 & (∀b ∈ I) fb a is finite-to-one .

Claim 1. I is a P-ideal.

Case 1: [X]ℵ0 ⊆ I0 for some uncountable X ⊆ S. Then

(∀b ∈ I) fb X is finite-to-one.

Case 2: S =⋃n<ω Sn such that Sn ⊥ I0 for all n < ω. Pick g : S → ω such that

(∀n) g−1(n) ⊆ Sn. Then g weakly coheres with every fb (b ∈ I).

44 STEVO TODORCEVIC

Corollary 22.4. Assume PID. Then for every family

fa : a→ ω (a ∈ I)

of weakly coherent functions indexed by a Pℵ1-ideal I on some set S there is

g : S → ω

which weakly coheres with every fa (a ∈ I).

Proof. Since I is a Pℵ1-ideal, the first alternative of the theorem is impossible.

Theorem 22.5 ([124]). PID implies that (κ) fails for all κ of cofinality > ω1.

Proof Sketch. We show using PID that no ordinal θ of cofinality > ω1 supports anontrivial coherent C-sequence Cα (α < θ).

Given a C-sequence Cα (α < θ), define

ρ2 : [θ]2 → ω

recursively byρ2(α, β) = ρ2 (α,min (Cβ \ α)) + 1,

with the initial condition ρ2(α, α) = 0. We shall need the following two propertiesof ρ2(see [128]):

(1) If Cα (α < θ) is coherent then

supξ<α|ρ2(ξ, α)− ρ2(ξ, β)| <∞

whenever α < β < θ.(2) If Cα (α < θ) is coherent and nontrivial then

(∀n < ω) (∀A,B ⊆ θ) [supA = supB = θ ⇒ (∃α ∈ A) (∃β ∈ B) ρ2(α, β) > n] .

Property (1) gives that

fα = ρ2(·, α) : α→ ω (α < θ)

is a family of weakly coherent mappings. Using PID, we have

g : θ → ω

such that for every α < θ,

(∀X ⊆ α) [fα X is unbounded ⇒ g X is unbounded] .

This is impossible if Cα (α < θ) has property (2).

It follows that PID has considerable large cardinal strength and that a deep workabout inner models of set theory and determinacy allows us to state the followingkind of consequences.

Theorem 22.6 ([55]). The P-ideal dichotomy implies the projective determinacy.

Definition 22.7. For functions

f : α→ ω and g : β → ω

defined on ordinals α and β respectively, set(1) f ≤l g if α < β and

(∀X ⊆ α) [f X is unbounded ⇒ g X is unbounded] .


(2) f ≤r g if α < β and

(∀X ⊆ α) [g X is unbounded ⇒ f X is unbounded] .

Note that above we have shown the following:

Theorem 22.8 ([124]). Assume PID. Then for every ordinal θ of cofinality > ω1

and every sequence fα : α→ ω (α < θ) such that

fα ≤l fβ whenever α < β,

there is g : θ → ω such that

fα ≤r g for all α < θ.

The following result shows that a similar fact holds for the ordering ≤r as well.

Theorem 22.9 ([135]). Assume PID. Then for every ordinal θ of cofinality > cand every sequence fα : α→ ω (α < θ) such that

fα ≤r fβ whenever α < β,

there is g : θ → ω such that

(∀X ∈ [θ]ℵ0) [g X is bounded ⇒ (∃α) fα X is bounded] .

Proof. Let

I = a ∈ [θ]ℵ0 : (∀α < θ)fα a is finite-to-one.

Claim 1. I is a P-ideal.

To see this, fix an : n < ω ⊆ I. If no b ⊆⋃n an such that b ⊇∗ an for all n

belongs to I then we can find β < θ such that for every such b there is α < β suchthat fα b is not finite-to-one. Since fα ≤l fβ for all α < β, it follows that fβ bis also not finite-to-one, a contradiction.

Corollary 22.10 ([135]). PID implies θℵ0 = θ for all regular cardinals θ ≥ c.

We prove this by induction on regular cardinals θ ≥ c. The only problematic case isa cardinal θ of the form θ = κ+ for some κ of cofinality ω. For this, fix a sequenceκn : n < ω of regular cardinals converging to κ and as before fix τ : [θ]2 → ωsuch that

τ(α, γ) ≤ max τ(α, β), τ(β, γ) for α < β < γ < θ(4)

|ξ < α : τ(ξ, α) ≤ n| ≤ κn for all α < θ and n < ω.(5)

Let fα := τ(·, α) : α→ ω (α < θ). Then fα ≤r fβ whenever α < β. Let g : θ → ωbe such that for every countable X ⊆ θ on which g is bounded there is α < θ suchthat fα is bounded on X. It follows that

θℵ0 = |⋃n g−1[n]| = |X ∈ [θ]ℵ0 : (∃α > supX) fα X is bounded

=∑α<θ,n∈ω |ξ < α : τ(ξ, α) ≤ n| = θ.

Corollary 22.11 ([135]). PID implies SCH.

46 STEVO TODORCEVIC

23. PID and the S-space problem

In this section we mention the original source of PID, a natural topologicalproblem asking whether every hereditarily separable regular space is Lindelof. Theimportance of this problem was fully established by the early 1960’s and it has beenfrequently referred as the S-space problem (see [87]). The following example pointsimmediately towards the nature of the problem and the fact that PID might berelevant.

Example 23.1 ([108]). If b = ω1 then there is a regular first countable hereditarilyseparable space which is not Lindelof. In fact the space is a subspace of NN wherethe usual separable metric topology is extended by making the sets of the formx : (∀n) x(n) ≤ y(n), (y ∈ NN) open.

Suppose X is a non-Lindelof subspace of some regular space K. This means thatfor each x ∈ X we can choose an open neighborhood Ux of x in K such that

X \⋃x∈X0

Ux 6= ∅ for all countable X0 ⊆ X.Since K is regular, for each x ∈ X we choose another open neighborhood Vx of xin K such that Vx ⊆ Ux. Let

IX = A ∈ [X]≤ℵ0 : (∀x ∈ X) A ∩ Vx ∈ FIN,the ideal of countable subsets of X that are orthogonal to Vx : x ∈ X. If b > |X|,the ideal IX is a P-ideal, so let us examine the two alternatives of PID. First of allnote that any Y ⊆ X with property [Y ]ℵ0 ⊆ IX must be discrete. On the otherhand if some subset Y of X is orthogonal to IX then it cannot contain a countablesubset D such that

(D \D) ∩⋃x∈X Ux = ∅,

where the closure of D is taken in K. So in particular if p > |X|, such a subset Yof X cannot be separable. This establishes the following.

Theorem 23.2 ([102]). Assuming PID, every regular hereditarily separable spaceof cardinality < p is Lindelof.52

So, in particular, assuming PID and p > ω1, we conclude that every regularhereditarily separable space is Lindelof, i.e. we have a solution to the S-spaceproblem. However, examining more closely the above argument one realizes thatthis argument is really about the ambient spaces K that have some compactnessas well as convergence properties leading us eventually to the following interestingresult.

Theorem 23.3 ([9], [41]). Assuming PFA, every compact countably tight space issequential.

Using a forcing and absoluteness argument this has the following fact53 as an im-mediate consequence.

Theorem 23.4. Assuming PFA, every compact countably tight space has a Gδ-point and is therefore sequentially compact.

52It is here that we were originally using the dual form of PID discussed above and saying that

for an ℵ1-generated ideal I on some uncountable set S either there is an uncountable subset of S

orthogonal to I or an uncountable subset X of S such that [X]ℵ0 ⊆ I.53This fact has been first observed by Alan Dow.


Looking at the analogues of these results in the context of descriptive set theoryone arrives at the following result whose proof however requires some Ramsey theory(see [129]; Ch. 7).

Theorem 23.5. Every countable analytic space54 with a countably tight compacti-fication is bisequential55.

Typical examples of countable analytic spaces are countable collections of Borelfunctions on some Polish, or more generally, completely metrizable space. Thefollowing well known result connects this with the theory of compact countablytight spaces.

Theorem 23.6 ([86]). Every compact set of Baire-class-1 functions defined onsome Polish space is countably tight and sequentially compact.

Combining Theorems 23.5 and 23.6 we get the following well-known result.

Theorem 23.7 ([14]). Every compact set of Baire-class-1 functions defined onsome Polish space is Frechet56.

Example 23.1 and Theorem 23.2 suggest the following natural question.

Question 23.8. Are any of the following statements equivalent under PID?(1) Every regular hereditarily separable space is Lindelof.(2) Every regular first countable hereditarily separable space is Lindelof.(3) Every compact hereditarily separable space is hereditarily Liindelof.(4) b = ω2.(5) p = ω2.

We finish this section by mentioning that the dual problem, the L-space problem,has a negative solution.

Theorem 23.9 ([77]). There is a regular hereditarily Lindelof nonseparable space.

24. PID and Classification of Transitive Relations on ω1

24.1. Tukey Reductions and Cofinal Types.

Definition 24.1. For two posets D and E, set

D ≤T E iff (∃f : D → E) [X is unbounded in D ⇒ f ′′X is unbounded in E.]

The map f : D → E witnessing D ≤T E is called a Tukey-map or Tukey-reduction.Put

D ≡T E iff D ≤T E and E ≤T D.Equivalence classes of ≡T are called Tukey types. We shall let D <T E denote thefact that D ≤T E but not E ≤T D.

Proposition 24.2 ([131]). The following are equivalent for two posets D and E :(1) D ≤T E,

54A topological space X is analytic if its topology is analytic as a subset of 2X .55Recall that a space X is bisequential if every ultrafilter U on X converging to a point x has

a subsequence An : n < ω ⊆ U converging to x.56Recall that a space X is Frechet if every subset of X which accumulates to a point of X

contains a sequence which converges to the point.

48 STEVO TODORCEVIC

(2) There is a convergent map g : E → D, i.e., a map with the property thatfor every d ∈ D there is e ∈ E such that g(e′) ≥ d for all e′ ≥ e,

(3) There is a cofinal map h : E → D, i.e., a map which maps cofinal subsetsof E to cofinal subsets of D

Remark 24.3. It is for this reason that sometimes Tukey-types are also calledcofinal types. Another reason is the following interesting result.

Theorem 24.4 ([131]). The following conditions are equivalent for every pair Dand E of directed sets:

(1) D ≡T E,(2) D and E can be isomorphically embedded as cofinal subsets of a third di-

rected poset X.

24.2. The Five Tukey Types.

[ω1]<ω

ω × ω1

@@@@

@@@@

@@@

ω ω1

1

~~~~~~~~~~~

???????????

Question 24.5 ([131]). Is there any other Tukey type below [ω1]<ω?

The first answer to this question given in [52] while incomplete did (at least im-plicitly) point towards the Tukey-classification theory of definable directed sets, anactive area of current research in set theory (see [42], [95]).

Theorem 24.6 ([52]). If CH holds there exist at least seven Tukey types of directedposets of cardinality at most ℵ1. For example, the lattices NN and `1 are pairwiseTukey-inequivalent and also not Tukey-equivalent to any of the five basic directedsets 1, ω, ω1, ω × ω1, and [ω1]<ω.

A more general construction can be obtained as follows. To a given topologicalspace X we associate a directed set as follows

K(X) = K ⊆ X : K compact.

It turns out that K(X) for X a subspace of ω1 (equipped with its natural ordertopology) realizes many different Tukey types of directed sets of cardinality contin-uum. In particular, we have the following fact showing that it is not reasonable toexpect that we can identify all cofinal types of directed sets of cardinality contin-uum.


Theorem 24.7 ([106]). For uncountable A,B ⊆ ω1, if there is a Tukey reductionfrom K(A) into K(B) then B \A must be non-stationary. So if for two uncountablesubsets A and B of ω1 have stationary symmetric difference, the correspondingdirected sets K(A) and K(B) are not Tukey-equivalent.

Example 24.8 ([107]). If b = ω1, there is a sublattice Db of NN such that

ω × ω1 <T Db <T [ω1]<ω .

For example, let Db be the sublattice of NN generated by an <∗-increasing <∗-unbounded sequence

fξ : ξ < b ⊆ N↑N.

Definition 24.9. A subset of D is strongly unbounded if it is infinite and if all ofits infinite subsets are unbounded.

Lemma 24.10. In any poset of cofinality < b the class of countable strongly un-bounded sets forms a P-ideal.

Proof. Consider a sequence An (n < ω) of countable strongly unbounded subsetsof D. Then for each d ∈ D there is fd ∈ NN such that

(⋃n<ω An(fd(n))) ∩ x ∈ D : x ≤D d = ∅,

where An(k) is our notation for the set of first k elements of An in some fixedenumeration of this set in type ω. Choose g ∈ NN such that d ∈ D : fd <∗ g iscofinal in D. Then B =

⋃n<ω An(g(n)) is a strongly unbounded subset of D such

that An ⊆∗ B for all n.

Definition 24.11. A subset of D is pseudo-bounded if it contains no stronglyunbounded subsets, or equivalently if every infinite subset of D contains an infinitebounded subset.

Theorem 24.12 ([106]). Assuming PID, for every directed (or undirected) posetD of cofinality < b, either

(1) [ω1[<ω≤T D, or(2) D can be decomposed into countably many pseudo-bounded subsets.

This suggests considering the class D0 of all directed sets D with the property thatpseudo-bounded countable subsets of D are in fact bounded.

Theorem 24.13 ([106]). The following statements are equivalent assuming PID:

(1) 1, ω, ω1, ω×ω1, and [ω1]<ω are all Tukey types of directed sets belonging tothe class D0 and having cofinality ≤ ω1;

(2) b = ω2.

Question 24.14. Assuming PID, is b = ω2 equivalent to the statement that

1, ω, ω1, ω × ω1, and [ω1]<ω

represent all the Tukey-types of directed sets of size ℵ1?

Lemma 24.15. In a directed set D of cofinality < p, every countable pseudo-bounded subset of D is in fact bounded.

50 STEVO TODORCEVIC

Proof. Consider a countable unbounded subset A of D and fix a cofinal subset D′

of D of cardinality < p. Then A \ x ∈ D : x ≤D d, (d ∈ D′) is a downwardsdirected family of infinite subsets of A of size < p so we can find infinite B ⊆ Awhich is almost included in every member of the family. It follows that B is stronglyunbounded in D and so in particular A is not pseudo-bounded.

Theorem 24.16 ([106]). Assuming PID and the equality p = ω2,

1, ω, ω1, ω × ω1, and [ω1]<ω

represent all Tukey-types of directed sets of cardinality at most ℵ1.

Question 24.17. Assuming PID, is p = ω2 equivalent to the statement that

1, ω, ω1, ω × ω1, and [ω1]<ω

represent all the Tukey-types of directed sets of size ℵ1?

A similar result holds for all posets of cardinality at most ℵ1 though the list ofpossible Tukey types is countably infinite. To succinctly state this result, let

D0 = 1, D1 = ω,D2 = ω1, D3 = ω × ω1 and D4 = [ω1]<ω1

be the ordered list of basic five directed posets of size at most ℵ1. We also needto recall that m is the minimal cardinal κ for which one can find a ccc poset Pand κ dense open subsets of P such that no filter of P meets all the dense sets,or equivalently m is the minimal cardinality of a ccc non-σ-centered poset. Thenm ≤ p ≤ b, so in the context of PID the cardinal m has only two possible values,m = ω1 or m = ω2. So in the context of PID, the statement m = ω2 is equivalentto MAℵ1 . We direct the reader to [109] and [112] for more information about m.

Theorem 24.18 ([117]). The following statements are equivalent assuming PID:(1) m = ω2.(2) Every poset of size ℵ1 is Tukey-equivalent to one from the list:

(a)⊕

i<5 niDi (i < 5, ni < ω),(b) ℵ0 · 1⊕

⊕4i=2 niDi (2 ≤ i < 5, ni < ω),

(c) ℵ0 · ω1 ⊕ n4[ω1]<ω (n4 < ω),(d) ℵ0 · [ω1]<ω,(e) ℵ1 · 1.

This leads us to the following question.

Question 24.19. Are any of the three statements b = ω2, p = ω2 and m = ω2

equivalent under PID?

25. PID and Weakly Distributive Boolean Algebras

Recall the following well known condition on a complete Boolean algebra.

Definition 25.1. A Boolean algebra B is weakly distributive if for every sequence

ank : n, k < ω ⊆ B,∧n

∨k

ank =∨F

∧n

anF (n),

where F ∈ Finω, and where for a finite set X ⊆ ω,

anX =∨k∈X

ank.


Particularly interesting are complete weakly distributive Boolean algebras satisfyingthe countable chain condition. The Souslin algebra, the algebra of regular-opensubsets of the ccc nowhere separable linearly ordered continuum, is such an algebra,a fact first pointed out in [71]. The existence of these algebras is easily seen to bein fact equivalent to the negation of Souslin hypothesis and so the following factrules out their existence inder PID.

Theorem 25.2 ([2]). The P-ideal dichotomy implies the Souslin hypothesis.

Proof. Given a Souslin tree T , let

IT =a ∈ [T ]≤ℵ0 : a is strongly unbounded in T

.

Claim 1. IT is a P-ideal (i.e. we don’t need to assume b > ω1).

Proof. Let an : n < ω ⊆ IT be given. Fix δ < ω1 such that⋃n<ω an ⊆ T δ and

let tk : k < ω = Tδ. Then for each k, n < ω, the set an(k) = s ∈ an : s <T tk isfinite. Let a =

⋃n<ω(an \

⋃k≤n an(k)). Then a ∈ IT and a ⊇∗ an for all n < ω.

Applying PID to IT , we have that either:(1) T contains an uncountable strongly unbounded set X. Impossible, because

X would in particular contain an uncountable antichain.(2) T can be covered with countably many pseudo-bounded subsets. Impossi-

ble, because a pseudo-bounded subset of T can be covered by finitely manychains of T . So in particular, T would have an uncountable chain.

Theorem 25.3 ([50]). PID implies that the Souslin hypothesis gets preserved underforcing by a measure algebra.

Proof. Fix a measure algebra R and an R-name ≤T such thatq(ω1, ≤T

)is a Souslin tree

y= 1

LetJ = a ∈ [ω1]≤ℵ0 : Ja ∈ IT K = 1.

Claim 1. J is a P-ideal.

Proof. Consider an : n < ω ⊆ J and fix δ < ω1 such thatq⋃

n<ω an ⊆ δy

= 1.

We assume thatq[δ, δ + ω) is the δth level

y= 1. Let ank be the R-name for the set

s ∈ an : s <T δ + k . For n,m < ω, find a finite set Fnm ⊆ an with the propertythat µ(J

⋃k≤n ank ⊆ FnmK) > 1− 2−m. Finally, let

a =⋃m<ω

(am \⋃n≤m

Fnm).

For m < ω, set am = a ∩⋃n≥m an. Then for all m ≥ k,

µ (Jpred(δ + k) ∩ am 6= ∅K) < 2−m+1

so we must have that Ja is strongly unboundedK = 1.

Applying PID, we have the following alternatives:

(1) There is uncountable X ⊆ ω1 such thatr

[X]ℵ0 ⊆ ITz

= 1. In this case,

J(ω1,≤T ) has an uncountable antichainK = 1.

52 STEVO TODORCEVIC

(2) There is uncountable Y ⊆ ω1 such that Y ⊥ J . In this case one shows thatsome condition of R forces that (ω1,≤T ) has an uncountable branch.

Definition 25.4. A subset X of some poset P is essentially strongly unbounded if

p ∈ P : x ∈ X : p ≤ x ∈ Finis dense-open in P.

Note 25.5. If P is a (normal) tree of height, say ω1, then essentially stronglyunbounded sets are strongly unbounded.

Proposition 25.6 ([82],[7]). If P is an ωω-bounding ccc poset then the ideal IP ofall countable essentially strongly unbounded subsets of P is a P-ideal.

Remark 25.7. If P = B \ 0 for some Boolean algebra B, the ωω-boundedness ofP is the same as the weak distributivity of the Boolean algebra B.

Definition 25.8. A function ν : B → [0,∞) is a strictly positive continuous sub-measure on B if

(1) ν(a) = 0 iff a = 0,(2) a ≤ b implies ν(a) ≤ ν(b),(3) ν(a ∨ b) ≤ ν(a) + ν(b),(4) an ↓ 0 implies ν(an)→ 0,

If (3) is replaced by a stronger condition(3)∗ ν(a ∨ b) = ν(a) + ν(b) whenever a ∧ b = 0,

then ν is a σ-additive measure (strictly positive as well).

The following problem is a special case of the problem asking for conditions thatwould guarantee the existence of strictly positive continuous submeasures on com-plete Boolean algebras (see, [8]).

Problem 25.9 ([72]57). Does every complete weakly distributive ccc algebra sup-port a strictly positive σ-additive measure?

This problems splits naturally into the following two parts second of which becameknown in the literature under the name of ’the control measure problem’.

Problem 25.10 ([71]).(1) Does every weakly distributive B satisfying the ccc support a strictly posi-

tive continuous submeasure?(2) Given that B supports a strictly positive continuous submeasure, does it

also support a strictly positive σ-additive measure?

Theorem 25.11 ([7]). The PID implies that every complete weakly distributivealgebra B satisfying the ccc supports a strictly positive continuous submeasure.

Proof Sketch. Apply PID to the P-ideal IB of all countable essentially stronglyunbounded subsets of B \ 0, which can also be viewed as the collection of

xn : n < ω ⊆ B+

such that ∧m

∨n≥m

xn = 0,

57Problem 163, put down originally by von Neumann on July 4, 1937.


i.e., sequences converging to 0. Note that since B is ccc there is no uncountableX ⊆ B \ 0 such that [X ]ℵ0 ⊆ IB and note that the second alternative of PIDmeans that 0 is a Gδ-point of B in the sequential order topology of B. Then use theresult of [6] to show that the sequential order topology of B is in fact metrizable.

Remark 25.12. More details about the Von Neumann and Maharam problemscan be found in articles [8] and [134]. We shall now work towards removing the useof PID from the analysis of this problem.

Definition 25.13 ([51]). A Boolean algebra satisfies the σ-finite chain conditionif it can be decomposed as

B =⋃n<ω

Bn

such that no Bn contains an infinite subset of pairwise disjoint elements.

Theorem 25.14 ([125]). The following are equivalent for every complete Booleanalgebra B:

(1) B carries a strictly positive continuous submeasure;(2) (a) B is weakly distributive, and

(b) B satisfies the σ-finite chain condition.

Remark 25.15. As the proof of [125] shows, the σ-finite chain condition in The-orem 25.14 can be replaced by any kind of countable chain condition which is pre-served by the proper poset (which could also be assumed not to add reals) forcingan instance of the P-ideal dichotomy. For example, the countable chain condition ofany complete Boolean algebra generated by an appropriately definable poset on aPolish space satisfies this requirement, so it will have a strictly positive continuousmeasure whenever it is weakly distributive (see [32]).

We finish this setion with a result which shows that the second part of Problem25.10 has a negative answer.

Theorem 25.16 ([100]). There is a complete Boolean algebra B carrying a strictlypositive continuous submeasure which supports no strictly positive σ-additive mea-sure.

26. PID and Biorthogonal Systems in Banach spaces

Recall that a biorthogonal system in some Banach space X is any sequence(xi, fi) : i ∈ I ⊆ X ×X∗ such that fi(xj) = δij , where δij is Kronecker’s delta.This concept has many variations and is related to many problems of this areaof functional analysis (see [46]). For example, the problem whether this conceptcharacterizes the separability of Banach spaces, i.e. whether every nonseparableBanach space has an uncountable biorthogonal system, has been around for quitesome time (see [19]). The following example points out towards the set-theoreticalnature of this question and hints that PID might be relevant.

Example 26.1 ([108]). If b = ω1 there is a compact nonmetrizable scattered spaceK such that its function space X = C(K) is hereditarily Lindelof relative to itsweak topology and so in particular X has no uncountable biorthogonal system.

It turns out that the right way to look at this problem is to work towards showingthat under certain assumptions the space X has a quotient with a Schauder basisof length ω1.

54 STEVO TODORCEVIC

Theorem 26.2 ([126]). Assuming PID, the following conditions are equivalent forevery Banach space X of density < p :

(1) X can be written as the union of a strictly increasing ω1-sequence of closedsubspaces.

(2) There is a bounded linear operator T : X → c0(ω1) with nonseparable range.

The condition (1) means that there is a sequence fξ (ξ < ω1) of norm-1 functionalssuch that

(∀ε > 0)(∀x ∈ X) ξ < ω1 : |fξ(x)| ≥ ε is countable.We get the condition (2) by finding an uncountable set Γ ⊆ ω1 such that

(∀ε > 0)(∀x ∈ X) ξ ∈ Γ : |fξ(x)| ≥ ε is finite.

To this end, let

I = A ∈ [ω1]≤ℵ0 : (∀ε > 0)(∀x ∈ X) ξ ∈ A : |fξ(x)| ≥ ε is finite.By our assumption on density of X, the ideal I is a P-ideal. Clearly, the firstalternative of PID is giving us the desired conclusion so we examine the second.Suppose Γ is an uncountable subset of ω1 orthogonal to I. Then again using ourassumption that p is bigger that the density of X, we conclude that, relative tothe weak*-topology of X∗, the set H = fξ : ξ ∈ Γ accumulates to 0∗ though nocountable subset of H does. Let

K =⋃α<ω1

fξ : ξ ∈ Γ ∩ α,where the closure is taken relative to the weak*-topology. Let

R = f − g : f, g ∈ K, f 6= g.We repeat the procedure by defining now the ideal

J = A ∈ [R]≤ℵ0 : (∀ε > 0)(∀x ∈ X)f ∈ A : |f(x)| ≥ ε is finite.Again this is a P-ideal and the first alternative of PID is giving the desired conclu-sion, so we finish the proof with the following claim.

Claim 1. The set R cannot be covered by countably many subsets orthogonal to J .

The relevance of condition (1) of Theorem 26.2 is explained by the following fact.

Theorem 26.3 ([126]). Assuming PID, the following conditions are equivalent forevery Banach space X of density < m :

(1) There is a bounded linear operator T : X → c0(ω1) with nonseparable range.(2) There is a quotient map Q : X → Y onto a Banach space Y with a mono-

tone Schauder basis of length ω1.

Corollary 26.4 ([126]). Assuming PID and m > ω1, every nonseparable Banachspace has an uncountable biorthogonal system.

The operators T : X → c0(ω1) with nonseparable range are usually given as

T (x) = (fξ(x))ξ<ω1 ,

where fξ (ξ < ω1) is a sequence of uniformly bounded functionals. The proof ofTheorem 26.3 gives us the quotient Y with a Schauder basis (eα)α<ω1 such that thecorresonding sequence (e∗α)α<ω1 of biorthogonal functionals is really a subsequenceof the given sequence fξ (ξ < ω1) which defines the operator T . It is for this reasonthat Theorem 26.3 (and its proof) has particularly strong influence on Banach


spaces of density ℵ1. For example, in this case the second alternative of Theorem26.3 will give us not only uncountable but also fundamental biorthogonal system(xα, gα) : α < ω1, i.e a biorthogonal system such that X = spanxα : α < ω1.In fact, going through the proof of a standard result in this area (Theorem 4.15of [46]), we see that the sequence (gα)α<ω1 can be chosen to be a subsequence of(fξ)ξ<ω1 as well. This gives us the following corollary to be used later.

Corollary 26.5. Assume PID and m > ω1. Let X be a Banach space of den-sity ℵ1 and that for some bounded sequence fξ : ξ < ω1 ⊆ X∗ the operatorT (x) = (fξ(x))ξ<ω1 maps X into a nonseparable subset of c0(ω1). Then X has afundamental biorthogonal system (xα, gα) : α < ω1 with (gα)α<ω1 a subsequenceof (fξ)ξ<ω1 .

We finish this section with two applications of Theorem 26.3. Recall that a pointx of closed convex subset C of X is its support point if there is f ∈ X∗ such that

f(x) = infy∈C < supy∈Cf(y).

In [85], Rolewicz investigated the existence of closed convex sets supported by allof its points. He noticed that no separable Banach space can contain such a convexsubset and after noting that many nonseparable Banach space do contain suchsets asked if this is true for all Banach spaces. It is known that some genericnonseparable Banach spaces fail to have convex subsets supported by all of theirpoints (see [12], [69], [66]).

Theorem 26.6 ([126]). Assuming PID and m > ω1, every nonseparable Banachspace contains a closed convex subset supported by all of its points.

Proof. Let (xi, fi) : i ∈ I be an uncountable biorthogonal system in some Banachspace X. Let C = convxi : i ∈ I, the closed convex hull of the points from thebiorthogonal system. Then it is easily checked that every point of C is its supportpoint.

Example 26.1 and Corollary 26.4 suggest the following natural question.

Question 26.7. Are the following statement equivalent under PID?

(1) Every nonseparable Banach space has an uncountable biorthogonal system.(2) b = ω2.

The second application is to a purely geometric problem first investigated byMazur [73] who proved that if a Banach space X has a Frechet differentiable normthen every closed convex subset of X is the intersection of closed balls of X (or,in short, the norm has the Mazur intersection property). The Mazur intersectionproperty is indeed closely related to the differentiability in the context of Banachspaces and for this reason is studied only in the context of spaces in which everyconvex continuous function is Frechet differentiable on a dense set of points, orequivalently in the class of Banach spaces X with the property that every separablesubspace of X has separable dual. This is a well studied class of Banach spacesknown in the literature as the class of strong differentiability spaces or Asplundspaces (see, for example, [28] and [20]). The close connection between the Mazurintersection property and biorthogonal systems is revealed by the following tworesults.

56 STEVO TODORCEVIC

Theorem 26.8. [56] Suppose that a Banach space X has a biorthogonal system(xi, fi) : i ∈ I ⊆ X × X∗ such that X∗ = spanfi : i ∈ I. Then X admits anequivalent norm with the Mazur intersection property.

Theorem 26.9. [56] Suppose that a nonseparable Banach space X admits an equiv-alent norm with the Mazur intersection property. Then for every ε > 0 there is anuncountable ε-biorthogonal system (xξ, fξ) : ξ < ω1 ⊆ X×X∗, i.e. a system suchthat fξ(xξ) = 1 and |fξ(xη)| ≤ ε for ξ 6= η.

So, in particular, the space X = C(K) of Example 26.1 does not admit renormingwith the Mazur intersection property and it is, therefore, natural to investigate therelevance of PID to the Mazur problem. This is done using the following resultwhich shows that the class of Asplund spaces of density ℵ1 satisfies the hypothesisof Corollary 26.5.

Lemma 26.10 ([5]). Suppose X is an Asplund space of density ℵ1 and with anuncountable biorthogonal system. Then there is a sequence yξ : ξ < ω1 ⊆ SX suchthat the operator f 7→ (f(yξ))ξ<ω1 maps X∗ into a nonseparable subset of c0(ω1).

Combining Corollaries 26.4 and 26.5, Theorem 26.8 and Lemma 26.10, we getthe following interesting solution to the Mazur problem.

Theorem 26.11 ([5]). Assuming PID, every Asplund space X of density < m ad-mits an equivalent norm such that every closed convex subset of X is the intersectionof closed balls relative to the new norm.

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60 STEVO TODORCEVIC

[103] Stevo Todorcevic, On a conjecture of R. Rado, Journal of the London Mathematical Society,

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[104] Stevo Todorcevic, A note on the proper forcing axiom. Axiomatic set theory (Boulder, Colo.,1983), 209–218, Contemp. Math., 31, Amer. Math. Soc., Providence, RI, 1984

[105] Stevo Todorcevic, Reflection of stationary sets. Circulated note, 1984.

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[113] Stevo Todorcevic, Two examples of Borel partially ordered sets with the countable chain

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[126] Stevo Todorcevic, Biorthogonal systems and quotient spaces via Baire category methods.

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[137] W. Hugh Woodin, The Axiom of Determinacy, Forcing Axioms, and the NonstationaryIdeal, De Gruyter, Berlin, New York, 1999.

Department of Mathematics, University of Toronto, Toronto, Ontario, Canada

M5S 2E4E-mail address: [email protected]

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